Sequences Series Part1

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Contents

Notes

Chapter 3 is all about patterns. Thus, even though STPM questions purely on sequences & series are rare compared to binomial expansion and AP & GP, this part is nevertheless very important.

Learning Objectives (Syllabus)

  • use explicit or a recursive formula for a sequence to find successive terms
  • determine whether a sequence is convergent or divergent and find the limit of a convergent sequence
  • use the \sum notation
  • use the method of differences to obtain the sum of a finite or a convergent infinite series

Prior Knowledge

  • general ability of recognizing patterns
  • AP & GP
  • partial fractions

Sequences

An easy way to describe sequence is a

  • set of numbers
  • on a certain order
  • with a certain rule

Examples

  • 1,3,5,\, 7,9,\ldots 1\xrightarrow{{\color{Blue}+2}}3\xrightarrow{{\color{Blue}+2}}5\xrightarrow{{\color{Blue}+2}}7\xrightarrow{{\color{Blue}+2}}9
  • 1,2,4,8,\, 16,32,\ldots 1\xrightarrow{{\color{Blue}\times2}}2\xrightarrow{{\color{Blue}\times2}}4\xrightarrow{{\color{Blue}\times2}}8\xrightarrow{{\color{Blue}\times2}}16\xrightarrow{{\color{Blue}\times2}}32
  • 2\cdot3,3\cdot4,4\cdot5, 5\cdot6,6\cdot7,\ldots {\color{Red}2}\cdot{\color{Blue}3},{\color{Red}3}\cdot{\color{Blue}4},{\color{Red}4}\cdot{\color{Blue}5},{\color{Red}5}\cdot{\color{Blue}6},{\color{Red}6}\cdot{\color{Blue}7}

Note : AP & GP are just special types of sequence. There are certainly many more types of sequences.

Notation

u_{n}\, to represent n^{th}\, term / general term. n\, normally starts from 1\, but can also sometimes start from 0\,.

Example

u_{n}=\frac{n}{n+1}, n \in \mathbb{N} Note :\mathbb{N}=\left\{ 1,2,3,\ldots \right\} This might not be written as it is understood.

  • u_{1} =\, =\frac{1}{2}\,
  • u_{2} =\, =\frac{2}{3}\,
  • u_{3} =\, =\frac{3}{4}\,
  • u_{10} =\, =\frac{10}{11}\,

Determining un

u_{1}\, u_{2}\, u_{3}\, u_{4}\, \ldots u_{n}\,
1,\, 2,\, 3,\, 4,\, \ldots,\, n\,
  • Note that\begin{array}{c} u_{{\color{Blue}1}} \\ {\color{Blue}1} \\ \end{array} \begin{array}{c} u_{{\color{Blue}2}} \\ {\color{Blue}2} \\ \end{array} \begin{array}{c} u_{{\color{Blue}3}} \\ {\color{Blue}3} \\ \end{array} \begin{array}{c} u_{{\color{Blue}4}} \\ {\color{Blue}4} \\ \end{array} \begin{array}{c} u_{{\color{Blue}n}} \\ {\color{Blue}??} \\ \end{array} \begin{array}{c} u_{{\color{Blue}n}} \\ {\color{Blue}n} \\ \end{array}
3,\, 4,\, 5,\, 6,\, \ldots,\, n+2\,
  • Compare \begin{array}{cccccc} 1 & 2 & 3 & 4 & \ldots & n \\ \big\downarrow & \big\downarrow & \big\downarrow & \big\downarrow & & \big\downarrow \\ 3 & 4 & 5 & 6 & & ??? \\ \end{array}
    • \begin{array}{cccccc} 1 & 2 & 3 & 4 & \ldots & n \\ \left\downarrow ^{{\color{Blue}+2}} \right. & \left\downarrow ^{{\color{Blue}+2}} \right. & \left\downarrow ^{{\color{Blue}+2}} \right. & \left\downarrow ^{{\color{Blue}+2}} \right. & & \left\downarrow ^{{\color{Blue}??}} \right. \\ 3 & 4 & 5 & 6 & & {\color{Blue}??} \\ \end{array} \begin{array}{c} n \\ \left\downarrow ^{{\color{Blue}+2}} \right. \\ n+2 \\ \end{array}
2,\, 4,\, 6,\, 8,\, \ldots,\, 2n\,
  • Compare \begin{array}{cccccc} 1 & 2 & 3 & 4 & \ldots & n \\ \big\downarrow & \big\downarrow &\big\downarrow & \big\downarrow & & \big\downarrow \\ 2 & 4 & 6 & 8 & & ??? \\ \end{array}
    • \begin{array}{ccccc} 1 & 2 & 3 & 4 & \ldots \\ \left\downarrow ^{{\color{Blue}\times 2}} \right. & \left\downarrow ^{{\color{Blue}\times 2}} \right. & \left\downarrow ^{{\color{Blue}\times 2}} \right. & \left\downarrow ^{{\color{Blue}\times 2}} \right. & \\ 2 & 4 & 6 & 8 & \\ \end{array} \begin{array}{c} n \\ \left\downarrow ^{{\color{Blue}\times 2}} \right. \\ 2n \\ \end{array}

3,\, 6,\, 9,\, 12,\, \ldots,\, 3n\,
  • \begin{array}{cccccc} 1 & 2 & 3 & 4 & \ldots & n \\ \left\downarrow ^{{\color{Blue}\times 3}} \right. & \left\downarrow ^{{\color{Blue}\times 3}} \right. & \left\downarrow ^{{\color{Blue}\times 3}} \right. & \left\downarrow ^{{\color{Blue}\times 3}} \right. & & \left\downarrow ^{{\color{Blue}\times 3}} \right. \\ 3 & 6 & 9 & 12 & & 3n \\ \end{array}
3,\, 5,\, 7,\, 9,\, \ldots,\, 2n+1\,
  • \begin{array}{cccccc} 1 & 2 & 3 & 4 & \ldots & n \\ 3 & 5 & 7 & 9 & & ?? \\ \end{array} Not as easy to see the pattern
  • However, if compare \begin{array}{cccccc} 2 & 4 & 6 & 8 & \ldots & 2n\\ \big\downarrow & \big\downarrow & \big\downarrow & \big\downarrow & & \big\downarrow \\ 3 & 5 & 7 & 9 & \ldots & ?? \end{array}
  • \begin{array}{cccccc} 2 & 4 & 6 & 8 & \ldots & 2n\\ \left\downarrow ^{{\color{Blue}+ 1}} \right. & \left\downarrow ^{{\color{Blue}+ 1}} \right. & \left\downarrow ^{{\color{Blue}+ 1}} \right. & \left\downarrow ^{{\color{Blue}+ 1}} \right. & & \left\downarrow ^{{\color{Blue}+ 1}} \right. \\ 3 & 5 & 7 & 9 & \ldots & 2n+1 \end{array}
  • But how do we know we compare those two? \begin{array}{c} 2 \xrightarrow{{\color{Blue}+2}} 4 \xrightarrow{{\color{Blue}+2}} 6 \xrightarrow{{\color{Blue}+2}} 8\\ 3 \xrightarrow{{\color{Blue}+2}} 5 \xrightarrow{{\color{Blue}+2}} 7 \xrightarrow{{\color{Blue}+2}} 9 \end{array}
  • Both will have 2n\, in the formula. In fact, every time we see a sequence have a pattern of +2\,, it will have 2n\, in the formula
  • We then "adjust" the formula, by adding/subtracting a number to "fit" the sequence
  • \begin{array}{c} 3 \xrightarrow{{\color{Blue}+2}} 5 \xrightarrow{{\color{Blue}+2}} 7 \xrightarrow{{\color{Blue}+2}} 9 \qquad {\color{Red}2n} \end{array} Check whether it "fits"
  • \begin{array}{ccccc} {\color{Blue}u_{1}} & u_{2} & u_{3} & u_{4} & \qquad u_{n}\\ \hline {\color{Blue}3} & 5 & 7 & 9 & \qquad {\color{Blue}2n} \end{array} u_{1}\, is going to be 2\,, we need it to be {\color{Blue}3}\,, thus we +1\,
  • \begin{array}{ccccc} {\color{Blue}u_{1}} & u_{2} & u_{3} & u_{4} & \qquad u_{n}\\ \hline {\color{Blue}3} & 5 & 7 & 9 & \qquad {\color{Blue}2n}+{\color{Red}1} \end{array} Check quickly that the formula also fits other terms to be sure (at least mentally check u_{3}\,): u_{2}=2\left(2\right)+1=5

1,\, 3,\, 5,\, 7,\, \ldots,\, 2n-1\,
  • Let try the above quick method
  • \begin{array}{c} 1 \rightarrow 3 \rightarrow 5 \rightarrow 7\end{array}
  • \begin{array}{c} 1 \xrightarrow{{\color{Red}+2}} 3 \xrightarrow{{\color{Red}+2}} 5 \xrightarrow{{\color{Red}+2}} 7 \end{array}
  • \begin{array}{c} {\color{Blue}1} \xrightarrow{+2} 3 \xrightarrow{+2} 5 \xrightarrow{+2} 7 \qquad {\color{Red}2n} \end{array} 2\left(1\right)={\color{Blue}2}
  • \begin{array}{c} 1 \xrightarrow{+2} 3 \xrightarrow{+2} 5 \xrightarrow{+2} 7 \qquad 2n{\color{Red}-1} \end{array} Checku_{2}=2\left(2\right)-1=3

1,\, 4,\, 7,\, 10,\, \ldots,\, 3n-2\,
  • 1 \xrightarrow{+3} 4 \xrightarrow{+3} 7 \xrightarrow{+3} 10 3n\, -2\, u_{2}=3\left(2\right)-2=4
2,\, 7,\, 12,\, 17,\, \ldots,\, 5n-3\,
  • If you find it too hard to do use above method, we can always fall back to using AP as an alternative
  • \underbrace{2 \xrightarrow{+5} 7 \xrightarrow{+5} 12 \xrightarrow{+5} 17}_{\mbox{AP ,}a=2,d=5}
  • u_{n}=2+\left(n-1\right)5=5n-3
  • Note that even with using AP formula, it pays to check that the formula does indeed "fit", in case of careless mistakes.

Note: As noted, the above are all AP's, thus we can always fall back to AP formula if needed. But the above method is much faster (there isn't even any need to write any working at all), and checking would have minimized the risk of careless mistakes.


u_{1}\, u_{2}\, u_{3}\, u_{4}\, \ldots u_{n}\,
2,\, 4,\, 8,\, 16,\, \ldots,\, 2^{n}\,
  • Note that 2 \xrightarrow{{\color{Blue}\times 2}}4\xrightarrow{{\color{Blue}\times 2}}8\xrightarrow{{\color{Blue}\times 2}}16
  • In fact, it can be written as\begin{array}{ccccc} 2 & 4 & 8 & 16 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ 2^{1} & 2^{2} & 2^{3} & 2^{4} \\ \end{array}
  • Hmm... \begin{array}{ccccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}} & \qquad u_{n}\\ \hline 2 & 4 & 8 & 16 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ 2^{{\color{Blue}1}} & 2^{{\color{Blue}2}} & 2^{{\color{Blue}3}} & 2^{{\color{Blue}4}} & \qquad {\color{Blue}??} \end{array} 2^{n}\,

4,\, 8,\, 16,\, 32,\, \ldots,\, 2^{n+1}\,
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}}\\ \hline 4 & 8 & 16 & 32 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ 2^{{\color{Blue}2}} & 2^{{\color{Blue}3}} & 2^{{\color{Blue}4}} & 2^{{\color{Blue}5}} \end{array} \begin{array}{c} u_{{\color{Blue}n}} \\ \hline 2^{{\color{Blue}n+1}} \end{array}

1,\, 2,\, 4,\, 8,\, \ldots,\, 2^{n-1}\,
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}}\\ \hline 1 & 2 & 4 & 8 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ 2^{{\color{Red}??}} & 2^{{\color{Blue}1}} & 2^{{\color{Blue}2}} & 2^{{\color{Blue}3}} \\ \end{array} \begin{array}{cccccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}} & \ldots & u_{{\color{Blue}n}}\\ \hline 1 & 2 & 4 & 8 & & \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow & & \\ 2^{{\color{Blue}0}} & 2^{{\color{Blue}1}} & 2^{{\color{Blue}2}} & 2^{{\color{Blue}3}}& & 2^{{\color{Blue}n-1}} \\ \end{array}

1,\, -3,\, 9,\, -27,\, \ldots,\, \left(-3\right)^{n-1}\,
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}}\\ \hline 1 & -3 & 9 & -27 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ \left(-3\right)^{{\color{Blue}0}} & \left(-3\right)^{{\color{Blue}1}} & \left(-3\right)^{{\color{Blue}2}} & \left(-3\right)^{{\color{Blue}3}} \\ \end{array} \begin{array}{c} u_{{\color{Blue}n}} \\ \hline \left(-3\right)^{{\color{Blue}n}} \end{array}
2,\, 6,\, 18,\, 54,\, \ldots,\, 2\left(3^{n-1}\right)
  • Probably best to use back GP
  • \underbrace{2 \xrightarrow{\times 3} 6 \xrightarrow{\times 3} 18 \xrightarrow{\times 3} 54}_{\mbox{GP ,}a=2,r=3}
  • u_{n}=2\left(3^{n-1}\right)

1,\, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots,\, \left(\frac{1}{2}\right)^{n-1}
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}}\\ \hline 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ \left(\frac{1}{2}\right)^{{\color{Blue}0}} & \left(\frac{1}{2}\right)^{{\color{Blue}1}} & \left(\frac{1}{2}\right)^{{\color{Blue}2}} & \left(\frac{1}{2}\right)^{{\color{Blue}3}} \\ \end{array} \begin{array}{c} u_{{\color{Blue}n}} \\ \hline \left(\frac{1}{2}\right)^{{\color{Blue}n-1}} \end{array}
\frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \frac{16}{81}, \ldots,\, \left(\frac{2}{3}\right)^{n}
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}}\\ \hline \frac{2}{3} & \frac{4}{9} & \frac{8}{27} & \frac{16}{81} \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ \left(\frac{2}{3}\right)^{{\color{Blue}1}} & \left(\frac{2}{3}\right)^{{\color{Blue}2}} & \left(\frac{2}{3}\right)^{{\color{Blue}3}} & \left(\frac{2}{3} \right)^{{\color{Blue}4}} \\ \end{array} \begin{array}{c} u_{{\color{Blue}n}} \\ \hline \left(\frac{2}{3}\right)^{{\color{Blue}n}} \end{array}

Note: The above are all GP's, thus we can always fall back to GP formula if needed. In fact, when the first term is not equal to common ratio, it's best to use GP.


u_{1}\, u_{2}\, u_{3}\, u_{4}\, \ldots u_{n}\,
1,\, 4,\, 9,\, 16,\, \ldots,\, n^{2}\,
  • 1\xrightarrow{+3}4\xrightarrow{+5}9\xrightarrow{+7}16
  • Neither an AP nor GP. But those numbers look very familiar.
  • \begin{array}{cccc} 1 & 4 & 9 & 16 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ \end{array} In fact, when we see this pattern (the numbers that are added form an AP), it is likely to have "square"
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}}\\ \hline 1 & 4 & 9 & 16 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow\\ {\color{Blue}1}^{2} & {\color{Blue}2}^{2} & {\color{Blue}3}^{2} & {\color{Blue}4}^{2}\\ \end{array} \begin{array}{c}u_{{\color{Blue}n}} \\\hline {\color{Blue}n}^{2}\end{array}

8,\, 27,\, 64,\, 125,\, \ldots,\, \left(n+1\right)^{3}\,
  • These numbers look familiar too
  • \begin{array}{cccc} 8 & 27 & 64 & 125 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow \\ 2^{3} & 3^{3} & 4^{3} & 5^{3} \\ \end{array}
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}}\\ \hline 8 & 27 & 64 & 125 \\ \big\updownarrow & \big\updownarrow & \big\updownarrow & \big\updownarrow\\ {\color{Blue}2}^{3} & {\color{Blue}3}^{3} & {\color{Blue}4}^{3} & {\color{Blue}5}^{3}\\ \end{array} \begin{array}{c}u_{{\color{Blue}n}} \\\hline \left({\color{Blue}n+1}\right)^{3}\end{array}


u_{1}\, u_{2}\, u_{3}\, u_{4}\, \ldots u_{n}\,
-1,\, 1,\, -1,\, 1,\, \ldots,\, \left(-1\right)^{n}\,
  • Hmm... where did we encounter this alternating positive negative pattern before? GP with negative common ratio, as r^{n}\, will have alternate between positive and negative if r\, is negative
  • Since the values here are only 1\, and -1\,, let's try \left(-1\right)^{n}, which is
    • -1\, when n\, is odd
    • 1\, when n\, is even
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Red}2}} & u_{{\color{Blue}3}} & u_{{\color{Red}4}}\\ \hline {\color{Blue}-1} & {\color{Red}1} & {\color{Blue}-1} & {\color{Red}1}\\ \end{array} Exactly what is needed

1,\, -1,\, 1,\, -1,\, \ldots,\, \left(-1\right)^{n+1}\,
  • \begin{array}{cccccc} u_{{\color{Blue}1}} & u_{{\color{Red}2}} & u_{{\color{Blue}3}} & u_{{\color{Red}4}} & \ldots & \\ \hline {\color{Blue}1} & {\color{Red}-1} & {\color{Blue}1} & {\color{Red}-1} & & \left(-1\right)^{n}\\ \end{array} Doesn't fit. How do we change the even and odd sequence?
  • \begin{array}{cccccc} u_{{\color{Blue}1}} & u_{{\color{Red}2}} & u_{{\color{Blue}3}} & u_{{\color{Red}4}} & \ldots & \\ \hline {\color{Blue}1} & {\color{Red}-1} & {\color{Blue}1} & {\color{Red}-1} & & \left(-1\right)^{n{\color{Red}+1}}\\ \end{array} Note that \left(-1\right)^{n{\color{Red}-1}} works fine too

1,\, -2,\, 3,\, -4,\, \ldots,\, \left(-1\right)^{n+1}n\,
  • It's not a GP, we can try to see the numbers and sign separately
  • \begin{array}{cccc} u_{{\color{Blue}1}} & u_{{\color{Blue}2}} & u_{{\color{Blue}3}} & u_{{\color{Blue}4}} \\ \hline {\color{Blue}1} & -{\color{Blue}2} & {\color{Blue}3} & -{\color{Blue}4}\\ \end{array} \begin{array}{c} u_{{\color{Blue}n}} \\ {\color{Blue}n}\\ \end{array}
  • \begin{array}{cccc} u_{{\color{Red}1}} & u_{{\color{Red}2}} & u_{{\color{Red}3}} & u_{{\color{Red}4}} \\ \hline {\color{Red}+}1 & {\color{Red}-}2 & {\color{Red}+}3 & {\color{Red}-}4\\ \end{array} \begin{array}{c} u_{n} \\ {\color{Red}\left(-1\right)^{n+1}}n\\ \end{array}

2,\, 0,\, 2,\, 0,\, \ldots,\, \left(-1\right)^{n+1}+1\,
  • This is harder to see by itself. However, if we compare
    • 2\xrightarrow{-2} 0 \xrightarrow{+2} 2 \xrightarrow{-2} 0
    • 1\xrightarrow{-2} -1 \xrightarrow{+2} 1 \xrightarrow{-2} -1
    • \begin{array}{cccccc} u_{1} & u_{2} & u_{3} & u_{4} &\ldots & u_{n} \\ \hline 1 & -1 & 1 & -1 & & \left(-1\right)^{n+1} \\ 2 & 0 & 2 & 0 & & \left(-1\right)^{n+1}+1 \\ \end{array}


u_{1}\, u_{2}\, u_{3}\, u_{4}\, \ldots u_{n}\,
\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots, \frac{n}{n+1}
  • Best to do each "part" separately. You will also see here it is the best if you can do directly the AP's and GP's using the quick method (but do remember to quickly check too).
  • \begin{array}{cccc} u_{1} & u_{2} & u_{3} & u_{4}\\ \hline \frac{{\color{Blue}1}}{2} & \frac{{\color{Blue}2}}{3} & \frac{{\color{Blue}3}}{4} & \frac{{\color{Blue}4}}{5} \end{array} \begin{array}{c} u_{n}\\ \hline \frac{{\color{Blue}n}}{} \end{array}
  • \begin{array}{cccc} u_{1} & u_{2} & u_{3} & u_{4}\\ \hline \frac{1}{{\color{Red}2}} & \frac{2}{{\color{Red}3}} & \frac{3}{{\color{Red}4}} & \frac{4}{{\color{Red}5}} \end{array} \begin{array}{c} u_{n}\\ \hline \frac{n}{{\color{Red}n+1}} \end{array}

1 \cdot 2, 2 \cdot 3, 3 \cdot 4, 4 \cdot 5, \ldots, n\left(n+1\right)
  • \begin{array}{cccc} u_{1} & u_{2} & u_{3} & u_{4}\\ \hline {\color{Blue}1}\cdot 2 & {\color{Blue}2}\cdot 3 & {\color{Blue}3}\cdot 4 & {\color{Blue}4}\cdot 5 \end{array} \begin{array}{c} u_{n}\\ \hline {\color{Blue}n} \end{array}
  • \begin{array}{cccc} u_{1} & u_{2} & u_{3} & u_{4}\\ \hline 1 \cdot {\color{Red}2} & 2 \cdot {\color{Red}3} & 3 \cdot {\color{Red}4} & 4 \cdot {\color{Red}5} \end{array} \begin{array}{c} u_{n}\\ \hline n{\left(\color{Red}n+1\right)} \end{array}

\frac{1}{2 \cdot 3}, \frac{2}{3 \cdot 4}, \frac{3}{4 \cdot 5}, \frac{4}{5 \cdot 6}, \ldots, \frac{n}{\left(n+1\right)\left(n+2\right)}
  • \begin{array}{cccc} \frac{{\color{Blue}1}}{2 \cdot 3} & \frac{{\color{Blue}2}}{3 \cdot 4} & \frac{{\color{Blue}3}}{4 \cdot 5} & \frac{{\color{Blue}4}}{5 \cdot 6} \end{array} \begin{array}{c} \frac{{\color{Blue}n}}{} \end{array}
  • \begin{array}{cccc} \frac{1}{{\color{Blue}2} \cdot 3} & \frac{2}{{\color{Blue}3} \cdot 4} & \frac{3}{{\color{Blue}4} \cdot 5} & \frac{4}{{\color{Blue}5} \cdot 6} \end{array} \begin{array}{c} \frac{n}{{\color{Blue}\left(n+1\right)}} \end{array}
  • \begin{array}{cccc} \frac{1}{ 2 \cdot {\color{Blue}3}} & \frac{2}{ 3 \cdot {\color{Blue}4}} & \frac{3}{ 4 \cdot {\color{Blue}5}} & \frac{4}{ 5 \cdot {\color{Blue}6}} \end{array} \begin{array}{c} \frac{n}{\left(n+1\right){\color{Blue}\left(n+2\right)}} \end{array}

\frac{2}{1 \cdot 3}, \frac{4}{3 \cdot 5}, \frac{6}{5 \cdot 7}, \frac{8}{7 \cdot 9}, \ldots, \frac{2n}{\left(2n-1\right)\left(2n+1\right)}

\begin{array}{cccccc} u_{1} & u_{2} & u_{3} & u_{4} & \ldots & u_{n}\\ \hline \frac{{\color{Blue}2}}{{\color{Red}1}\cdot {\color{Purple}3}} & \frac{{\color{Blue}3}}{{\color{Red}3}\cdot {\color{Purple}5}} & \frac{{\color{Blue}4}}{{\color{Red}5}\cdot {\color{Purple}7}} & \frac{{\color{Blue}8}}{{\color{Red}7}\cdot {\color{Purple}9}} & & \frac{{\color{Blue}2n}}{{\color{Red}\left(2n-1\right)}\cdot {\color{Purple}\left(2n+1\right)}} \end{array}

\frac{2}{1 \cdot 3}, \frac{5}{4 \cdot 5}, \frac{8}{7 \cdot 7}, \frac{11}{10 \cdot 9}, \ldots, \frac{3n-1}{\left(3n-2\right)\left(2n+1\right)}

\begin{array}{cccccc} u_{1} & u_{2} & u_{3} & u_{4} & \ldots & u_{n}\\ \hline \frac{{\color{Blue}2}}{{\color{Red}1}\cdot {\color{Purple}3}} & \frac{{\color{Blue}5}}{{\color{Red}4}\cdot {\color{Purple}5}} & \frac{{\color{Blue}8}}{{\color{Red}7}\cdot {\color{Purple}7}} & \frac{{\color{Blue}11}}{{\color{Red}10}\cdot {\color{Purple}9}} & & \frac{{\color{Blue}3n-1}}{{\color{Red}\left(3n-2\right)}\cdot {\color{Purple}\left(2n+1\right)}} \end{array}

\frac{1^{2}}{3}, \frac{2^{2}}{7}, \frac{3^{2}}{11}, \frac{4^{2}}{15}, \ldots, \frac{n^{2}}{4n-1}

\begin{array}{cccccc} u_{1} & u_{2} & u_{3} & u_{4} & \ldots & u_{n}\\ \hline \frac{{\color{Blue}1}^{2}}{{\color{Red}3}} & \frac{{\color{Blue}2}^{2}}{{\color{Red}17}} & \frac{{\color{Blue}3}^{2}}{{\color{Red}11}} & \frac{{\color{Blue}4}^{2}}{{\color{Red}15}} & & \frac{{\color{Blue}n}^{2}}{{\color{Red}4n-1}} \end{array}

1+\frac{1}{2}, 1+\frac{1}{3}, 1+\frac{1}{4}, 1+\frac{1}{5}, \ldots, 1+\frac{1}{n+1}

\begin{array}{cccccc} u_{1} & u_{2} & u_{3} & u_{4} & \ldots & u_{n}\\ \hline 1+\frac{1}{{\color{Blue}2}} & 1+\frac{1}{{\color{Blue}3}} & 1+\frac{1}{{\color{Blue}4}} & 1+\frac{1}{{\color{Blue}5}} & & 1+\frac{1}{{\color{Blue}n+1}} \end{array}

Recursive Formula

1,1,2,3,5,8,\, Can you guess the next term?

  • 13,\; 21,\; 34,\ldots\;
  • Still can't see?
    • \overbrace{{\color{Blue}1,1}}^{{\color{Blue}+}}, {\color{Red}2}
    • 1\overbrace{{\color{Blue}1,2}}^{{\color{Blue}+}}, {\color{Red}3}
    • 1,1\overbrace{{\color{Blue}2,3}}^{{\color{Blue}+}}, {\color{Red}5}
    • 1,1,2\overbrace{{\color{Blue}3,5}}^{{\color{Blue}+}}, {\color{Red}8}
    • 1,1,2,3\overbrace{{\color{Blue}5,8}}^{{\color{Blue}+}}, {\color{Red}13}

  • So is this a sequence? Yes. As long as there is an order and a rule, it is a sequence, even though it's not easy to find a simple u_{n}\, formula.
  • So how would we describe this sequence with a formula?
    • To get the third term, we added the first and the second. u_{3}=u_{2}+u_{1}\,
    • The fourth them is the sum of the second and third term. u_{4}=u_{3}+u_{2}\,
    • u_{5}=\, u_{4}+u_{3}\,
    • So we have \begin{array}{ccccc} u_{3} & = & u_{2} & + & u_{1}\\ u_{4} & = & u_{3} & + & u_{2}\\ u_{5} & = & u_{4} & + & u_{3}\\ \ldots \\ u_{{\color{Blue}???}} & = & u_{{\color{Blue}??}} & + & u_{{\color{Blue}?}}\\ \end{array}
    • We know that{\color{Blue}???}\xleftarrow{+1}{\color{Blue}??}\xleftarrow{+1}{\color{Blue}?}, thus, to avoid minus, we let the smallest one be n\,
    • u_{{\color{Blue}n+2}}=u_{{\color{Blue}n+1}}+u_{{\color{Blue}n}}
    • But this formula only starts from u_{3}\,, the first two terms have to be noted down by itself
    • u_{n+2}=u_{n+1}+u_{n},\; u_{1}=1,\; u_{2}=1\,

Examples

Write the next 3 terms of the sequence

  • i) u_{n+1}=3u_{n}-1,\;u_{1}=1
    • u_{2}\,
      • =3u_{1}-1\, Note that \begin{align} u_{{\color{Blue}n+1}}&=3u_{{\color{Red}n}}-1 \\ u_{{\color{Blue}2}} & =3u_{{\color{Red}1}}-1 \end{align}
      • =3\left(1\right)-1=2
    • u_{3}\, =3u_{2}-1\, =3\left(2\right)-1=5
    • u_{4}\, =3u_{3}-1\, =3\left(5\right)-1=14

  • ii) u_{n+2}=\left(u_{n+1}\right)\left(u_{n}\right),\;u_{1}=1,\;u_{2}=2
    • u_{3}\, =\left(u_{2}\right)\left(u_{1}\right) =\left(2\right)\left(1\right)=2
    • u_{4}\, =\left(u_{3}\right)\left(u_{2}\right) =\left(2\right)\left(2\right)=4
    • u_{5}\, =\left(u_{4}\right)\left(u_{3}\right) =\left(2\right)\left(4\right)=8

Limits to Infinity

Example 1 :u_{n}=\frac{1}{n}

  • The actual terms will be \frac{1}{1},\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5},\ldots
  • Let's study the values of the terms
    • \begin{array}{ccccccccccc} \frac{1}{1} & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5}\\ \hline 1 & 0.5 & 0.333\ldots & 0.25 & 0.20 \end{array}
    • We can see the values are becoming smaller, as the value of n\, increases
    • In fact, if we continue, \begin{array}{cccccc} \ldots & \frac{1}{10} & \ldots & \frac{1}{100} & \ldots & \frac{1}{100000}\\ \hline \ldots & 0.1 & \ldots & 0.01 & \ldots & 0.00001\\ \end{array}
    • we can see the values are getting smaller and smaller and getting nearer and nearer to 0\,

  • In short, as n\, increases,
    • the value of \frac{1}{n}\, decreases
    • and approaches 0\,
  • Still not short enough, so we use limit notation
    • \lim represents limit
    • \lim{\color{Red}\frac{1}{n}} limit of \frac{1}{n}
    • \lim_{{\color{Red}n}\quad}\frac{1}{n} limit of \frac{1}{n} as n\,
    • \lim_{n\to{\color{Red}\infty}}\frac{1}{n} limit of \frac{1}{n} as n\, approaches infinity (meaning as n\, increases indefinitely)
    • Thus, we say \lim_{n\to \infty}\frac{1}{n} =0\,


Example 2 :u_{n}=\frac{n}{n+1}

  • \frac{1}{2},\frac{2}{3}, \frac{3}{4}, \frac{4}{5},\ldots
  • \begin{array}{cccccccccccccccc} \frac{1}{2} & \frac{2}{3} & \frac{3}{4} & \frac{4}{5} & \ldots & \frac{9}{10} & \ldots & \frac{99}{100} & \ldots & \frac{99999}{100000}\\ \hline 0.5 & 0.666\ldots & 0.75 & 0.8 & \ldots & 0.9 & \ldots & 0.99 & \ldots & 0.99999\\ \end{array} (we just need use calculator/mental calculations without writing all these down)
  • As n\, increases, the value of \frac{n}{n+1}\, increases and approaches 1\,
  • \lim_{n\to \infty}\frac{n}{n+1} =1\,


Example 3 :u_{n}=n^{2}\,

  • 1,4,9,16,25,\ldots
  • The values increases, and it keeps on increasing indefinitely large 10^{2}=100, 100^{2}=10000, 100000^{2}=1 \times 10^{10}
  • So what is its limit? There isn't any.
  • \lim_{n\to \infty}n^{2} does not exist
  • Alternatively, we can write \lim_{n\to \infty}n^{2}=\infty
    • but it sill doesn't mean the limit exist
    • seems a bit weird to say it doesn't exist when we actually write "=\,", but \lim_{n\to \infty}n^{2}=\infty simply means it becomes indefinitely large, which also means limit doesn't exist

Finding Limits to Infinity

Let's study the behavior of some common forms of u_{n}\,

As n\to \infty

  • 2n\, \to \infty
    • n\, becomes infinitely large, 2n\, also becomes infinitely large. Note that there is no meaning (and wrong) to write 2\infty
      • In other words, if somethings is large without limit, if we double it, it is still large without limit.

  • n+1\, \to \infty
    • it is likewise meaningless to write \infty + 1
  • n-100\, \to \infty
    • Note that even though it starts with -99, -98, -97 \ldots, sooner or later it will reach zero, and then positive, and keep on increasing infinitely large
  • -0.1n\, \to -\infty
    • \begin{align} & -0.1\left(10\right)={\color{Blue}-1} \\ & -0.1\left(100\right)={\color{Blue}-10}\\ & -0.1\left(10000\right)={\color{Blue}-1000}\\ \end{align} it becomes infinitely negative
  • 100-n\, \to -\infty
    • Even if it starts from 99\,, sooner or later it will be negative and keep on decreasing infinitely negative


  • n\, \to \infty
  • n^{2}\, \to \infty In fact, n^{2}\, increases faster than n\,
  • n^{4}\, \to \infty
  • \sqrt{n} \to \infty
    • \begin{align} & \sqrt{100}={\color{Blue}10} \\ & \sqrt{10000}={\color{Blue}100} \\ & \sqrt{1000000}={\color{Blue}1000} \\ \end{align} Note that even though \sqrt{n} increases much slower than n\,, it is still increasing indefinitely ( we can make n\, as large as we like by choosing suitable n\,)

Conclusion : We can see that if the term have n\, or positive powers of n\, it will likely not to have a limit ( which makes sense as n \to \infty ). Any fixed number, however large, is sooner or later going to be negligibly small compared to a very large n\,


As n\to \infty

  • \frac{1}{n} \to 0
    • As {\color{Blue}n} becomes larger, \frac{1}{{\color{Blue}n}} becomes smaller
      • 1\, divided by an increasingly large umber will become smaller and smaller
  • \frac{1}{\sqrt{n}} \to 0 (Since \sqrt{n}\to \infty)
  • \frac{1}{n+1} \to 0 (Since n+1\to \infty)
  • \frac{300}{n^{2}+3} \to 0
    • 300\, will be negligibly small compared to n^{2}\, as n\to \infty
    • In fact, even putting a million or billion will not change the fact that a fixed number will decrease approaching zero when divided by an infinitely increasing value
  • \frac{1}{n\left(n-10\right)} \to 0 (n\left(n-10\right)\to \infty)
  • \frac{1}{n!} \to 0 (in fact, n!\, increases very very fast, 70!\, will already result in error in calculator)

Conclusion : We can see that if the term have n\, or positive powers of n\, in the denominator only it will likely to have a limit of zero. Any fixed number, however large, is sooner or later going to be negligibly small when divided by an infinitely increasing value of n\, or positive powers or it.

Examples

  • \lim_{n\to \infty}n^{2}-10 \mbox{does not exist}\,
  • \lim_{n\to \infty}\frac{10}{3n-1} =0\,
  • \lim_{n\to \infty}3+\frac{1}{n} =3\,
    • We know that \frac{1}{n}\to 0 Thus, we just note it down like this \lim_{n\to \infty}3+{\color{Red}\cancelto{0}{\frac{1}{n}}} =3\,
  • \lim_{n\to \infty}n+\frac{4}{n^{2}} \mbox{does not exist}\,
    • Note that even though \frac{4}{n^{2}}\to 0, \lim_{n\to \infty}n does not exist

Rearranging

However, what would happen if we have n\, in the numerator and also denominator? What will happen when we divide an indefinitely increasing value with another indefinitely increasing value? One of the way we can do it is this:

  • If we have n\,(or higher degrees) on top and below, divide ALL terms by highest degree of n\,

Examples

  • \lim_{n\to \infty}\frac{n}{n+1}
    • First, note that \lim_{n\to \infty}\frac{{\color{Blue}n}}{{\color{Red}n}+1}(at numerator and also denominator), thus it's not as easy (though possible) to make direct conclusion.
    • Instead, let's try the above method and see how and why it works
      • Identify the highest degree \lim_{n\to \infty}\frac{{\color{Blue}n}}{{\color{Blue}n}+1} {\color{Blue}n}
      • Divide all terms by highest degree
        • Note that we actually divide top and bottom by n\,, but \lim_{n\to \infty}\frac{n{\color{Red}{\div n}}}{\left(n+1\right){\color{Red}{\div n}}} also means \lim_{n\to \infty}\frac{n{\color{Red}^{\div n}}}{n{\color{Red}^{\div n}}+1{\color{Red}^{\div n}}}
        • Thus \lim_{n\to \infty}\frac{n}{n+1}=
          • \lim_{n\to \infty}\frac{{\color{Red}1}}{\qquad} since \frac{n}{n}=1
          • \lim_{n\to \infty}\frac{1}{{\color{Red}1}\qquad}
          • \lim_{n\to \infty}\frac{1}{1+{\color{Red}\frac{1}{n}}}
          • Note also that we are just rearranging it, the \lim_{n\to \infty} is still there
      • Solve
        • \lim_{n\to \infty}\frac{n}{n+1}=\lim_{n\to \infty}\frac{1}{1+\frac{1}{n}} \lim_{n\to \infty}\frac{1}{1+{\color{Red}\cancelto{0}\frac{1}{n}}}
          • =1\,
    • Why does it work? When we divide with n\,, all the terms will become numbers or number divided by n\,, both which is easy to deduce it's limit
    • Is there a way we can see the limit directly? Actually, yes. Note that as n\to \infty, the \frac{n}{n+{\color{Red}1}} becomes negligibly small compared to n\,, and thus, doesn't really affect the value much, and we end up with \approx \frac{n}{n}=1. We can also see from the values \frac{10}{1{\color{Red}1}}\ldots \frac{100}{10{\color{Red}1}} \ldots \frac{100000}{10000{\color{Red}1}}
  • \lim_{n\to \infty}\frac{2n-10}{n+3}
  • \lim_{n\to \infty}\frac{2n-10}{n+3}
    • =\lim_{n\to \infty}\frac{2-\cancelto{0}{\frac{10}{n}}}{1+\cancelto{0}{\frac{3}{n}}} Note that we will "cancel" in this step, and then directly on the next step, there is no more "\lim"
    • =2\,
  • \lim_{n\to \infty}\frac{3n\left(n+2\right)}{5\left(n-1\right)\left(n+1\right)}
    • First we would need to expand (DO NOT divide terms before expanding)
      • \lim_{n\to \infty}\frac{3n\left(n+2\right)}{5\left(n-1\right)\left(n+1\right)}=\lim_{n\to \infty}\frac{3n^{2}+6n}{5n^{2}-5}
    • The highest degree is n^{2}\,
      • =\lim_{n\to \infty}\frac{3+\frac{6}{n}}{5-\frac{5}{n^{2}}} \lim_{n\to \infty}\frac{3+\cancelto{0}{\frac{6}{n}}}{5-\cancelto{0}{\frac{5}{n^{2}}}}
    • =\frac{3}{5}
    • We can see that we need to divide by highest degree (n^{2}\,) so that the highest term will be reduced to a number. It will not work if we have divided by n\,.
    • We can also see \lim_{n\to \infty}\frac{{\color{Blue}3}n^{2}+{\color{Red}6n}}{{\color{Blue}5}n^{2}{\color{Red}-5}}, that both -5\, and even 6n\, will become negligibly small compared to the highest power n^{2}\, (even if n\, is a million, n^{2}\, will be million squared, which will make n\, very small in comparison), which is why the coefficients of those terms will be the limit
  • \lim_{n\to \infty}\frac{n}{n^{2}-3}
  • \lim_{n\to \infty}\frac{n}{n^{2}-3}=\lim_{n\to \infty}\frac{\cancelto{0}{\frac{1}{n}}}{1-\cancelto{0}{\frac{3}{n^{2}}}} =0\,
    • We can also just compare the degrees \frac{{\color{Blue}n}}{{\color{Red}n^{2}}-3} Since the degree on the denominator is higher, the limit will be zero
  • \lim_{n\to \infty}\frac{n^{2}}{10n+3}
  • \lim_{n\to \infty}\frac{n^{2}}{10n+3}=\lim_{n\to \infty}\frac{1}{\cancelto{0}{\frac{10}{n}}+\cancelto{0}{\frac{3}{n^{2}}}} \mbox{does not exist}\,
    • Note that \frac{1}{x}\to \infty if x \to 0
    • It will be better to compare degrees in this case, or at least use comparing degrees to check (since it will be so easy to make a mistake)
      • \frac{{\color{Blue}n^{2}}}{10{\color{Red}n}+3} Since the degree on numerator is higher, the limit does not exist

rn (GP)

As n\to \infty(you can use calculator to investigate, put in increasingly large values of n\, and observe what happens)

  • 2^{n}\, \to \infty
  • \left(0.5\right)^{n} \to 0 Note that numbers such as (as shown by calculator)\left(0.5\right)^{100}=7.8886009051 \times 10^{{\color{Red}-31}} is very small
  • \left(1.1\right)^{n} \to \infty \left(1.1\right)^{1000}=2.469932918 \times 10^{{\color{Red}41}} is very large
  • \left(0.9\right)^{n} \to 0
  • \left(-2\right)^{n} \to \pm\infty The values alternate between large positive values and very negative values
  • \left(-0.3\right)^{n} \to 0 \left(-0.3\right)^{99}=-1.717925069 \times 10^{{\color{Red}-52}} is very small negative number (near zero)

Do keep in mind that this is because, when you multiply with

  • 1\, the value stay the same
  • a number greater than 1\, the value increases, so if you multiply it many times (r^{n}=r\times r \times r \times r \times \ldots ), it will become infinitely large (even if it is just slightly larger than 1\,)
  • a number between 0\, and 1\, the value decreases, so if you multiply it many times it will become smaller and smaller approaching zero (even if it is just slightly smaller than 1\,)
  • We see from above that it is similar for negative numbers.

Thus,

  • If \left|r\right|<1, \lim_{n\to \infty} r^{n} =0\,
  • If \left|r\right|>1, \lim_{n\to \infty} r^{n} \mbox{does not exist}\,

Example

  • \lim_{n\to \infty} 3^{n} \mbox{does not exist}\,
  • \lim_{n\to \infty} \left(\frac{2}{3}\right)^{n-1} =0\, \left|\frac{2}{3}\right|<1,, and \left(\frac{2}{3}\right)^{n{\color{Red}-1}} doesn't affect the fact that it will decrease approaching zero
  • \lim_{n\to \infty} \left(\frac{5}{4}\right)^{2n} \mbox{does not exist}\, \left|\frac{5}{4}\right|>1,, do not assume all fractions are smaller than zero!

Comparison

As n \to \infty

  • positive powers of n\, limit does not exist
  • constant divided by positive powers of n\, limit is zero
  • equal powers in numerator and denominator limit exist (divide by highest degree to find)
  • degree of denominator higher than numerator limit is zero (still can use above method but use this to check)
  • degree of numerator higher than denominator limit does not exist (still can use above method but use this to check)
  • r^{n}, \left|r\right|<1 limit is zero
  • r^{n}, \left|r\right|>1 limit does not exist

Other Examples

  • \lim_{n\to \infty}\frac{n^{2}}{3^{n}} =0\,
  • Both numerator and denominator are increasing, and it's not easy to compare them. The best way is just to use calculator and enter increasingly large value of n\,
  • \lim_{n\to \infty}3 =3\,
  • This can be seen by writing out the sequence : 3,3,3,3,3,\ldots
  • \lim_{n\to \infty}\left(-1\right)^{n} \,\mbox{does not exist}
  • This can be seen by writing out the sequence : -1,1,-1,1,-1,\ldots Note that alternating between two values cannot be considered as having a limit

Convergent / Divergent sequence

  • A sequence is said to be convergent if \lim_{n\to\infty}u_{n} exists
  • A sequence is said to be divergent if \lim_{n\to\infty}u_{n} does not exist

Note

  • The limit just need to exist (to be convergent), it does not matter what the value it is (does not need to be zero)
  • We can write limit =\infty (for cases where it is so), but it still means it does not exist, which is the only thing that matters to determine whether it is convergent or divergent.
  • Though it it usually helpful to try to understand the "why" of a definition from a logical point of view, in this case, just stick to the definition, and not try any other method to determine whether it is convergent or divergent, as it won't be accepted.

Examples

Determine if each of the following sequences are convergent if divergent. For those that are convergent, find their limits

  • \frac{2}{1}, \frac{3}{2}, \frac{4}{3}, \frac{5}{4},\ldots
    • Firstly, note again that we must use the definition. That is, even though we can see the terms are "converging" (getting nearer and nearer) to the value of 1\,, we still must use the definition to prove it
      • Before we can find \lim_{n\to \infty}u_{n}, we would need to find u_{n}\,
    • u_{n}=\, \frac{n+1}{n}
    • \lim_{n\to \infty}u_{n} Always start with this line and then only substitute u_{n}\,
      • =\lim_{n\to \infty}\frac{n+1}{n} In this case, its easier just to separate the terms
      • =\lim_{n\to \infty}1+\cancelto{0}{\frac{1}{n}}
      • =1\, We can see the limit exist
    • \therefore \mbox{The sequence is} \mbox{convergent}\,
    • \mbox{limit }=1\, Question did ask us to state the limit, so just state it again. it might seem a bit weird that the question asks us to determine convergence/divergence, and then ask us state the limit, but in our answer, we did it the other way round, but this it is the correct way.
  • 1.5, \left(1.5\right)^{2}, \left(1.5\right)^{3}, \left(1.5\right)^{4},\ldots
    • So the steps are
      • Determine u_{n}\,
      • Determine whether \lim_{n\to\infty}u_{n}\, exist or not
      • Determine whether sequence is convergent or divergent
    • u_{n}\, =\left(1.5\right)^{n}
    • \lim_{n\to\infty}u_{n} =\lim_{n\to\infty}\left(1.5\right)^{n} \mbox{does not exist}\, Note that although its not wrong to write \lim_{n\to\infty}\left(1.5\right)^{n}=\infty, there is no reason to do so here (and it might only confuse us about the existence of a limit)
    • \therefore \mbox{The sequence is} \mbox{divergent}\,

Exercise 1

For each of the following, determine u_{n}\,. Hence determine if the sequence are convergent or divergent. For those that are convergent, find their limit

1) \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\ldots u_{n}=\frac{n}{n+1},\mbox{convergent}, \mbox{limit} =1

  • \begin{align} & u_{n}=\frac{n}{n+1} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\frac{n}{n+1} \\ & \qquad \quad \; = \lim_{n \to \infty} \frac{1}{1+\cancelto{0}{\frac{1}{n}}}\\ & \qquad \quad \; = 1\\ & \therefore \mbox{Sequence is convergent, }\mbox{limit } =1 \end{align}

2) \frac{2}{1},\frac{2}{3},\frac{2}{5},\frac{2}{7},\ldots u_{n}=\frac{2}{2n-1},\mbox{convergent}, \mbox{limit} =0

  • \begin{align} & u_{n}=\frac{2}{2n-1} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\frac{2}{2n-1} \\ & \qquad \quad \; = 0\\ & \therefore \mbox{Sequence is convergent, }\mbox{limit } =0 \end{align}

3) \frac{1}{1},\frac{3}{4},\frac{5}{7},\frac{7}{10},\ldots u_{n}=\frac{2n-1}{3n-2},\mbox{convergent}, \mbox{limit} =\frac{2}{3}

  • \begin{align} & u_{n}=\frac{2n-1}{3n-2} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\frac{2n-1}{3n-2} \\ & \qquad \quad \; = \lim_{n \to \infty} \frac{2-\cancelto{0}{\frac{1}{n}}}{3-\cancelto{0}{\frac{2}{n}}}\\ & \qquad \quad \; = \frac{2}{3}\\ & \therefore \mbox{Sequence is convergent, }\mbox{limit } =\frac{2}{3} \end{align}

4) \frac{1^{2}}{3},\frac{2^{2}}{6},\frac{3^{2}}{9},\frac{4^{2}}{12},\ldots u_{n}=\frac{n}{3},\mbox{divergent}

  • \begin{align} & u_{n}=\frac{n^{2}}{3n}=\frac{n}{3} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\frac{n}{3} \mbox{ does not exist} \\ & \therefore \mbox{Sequence is divergent} \end{align}

5) \frac{1}{2\cdot 3},\frac{2}{3\cdot 4},\frac{3}{4\cdot 5},\frac{4}{5\cdot 6},\ldots u_{n}=\frac{n}{\left(n+1\right)\left(n+2\right)},\mbox{convergent}, \mbox{limit} =0

  • \begin{align} & u_{n}=\frac{n}{n+1} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\frac{n}{\left(n+1\right)\left(n+2\right)} \\ & \qquad \quad \; = \lim_{n \to \infty}\frac{n}{n^{2}+3n+2} \\ & \qquad \quad \; = \lim_{n \to \infty}\frac{\cancelto{0}{\frac{1}{n}}}{1+\cancelto{0}{\frac{3}{n}}+\cancelto{0}{\frac{2}{n^{2}}}} \\ & \qquad \quad \; = 0\\ & \therefore \mbox{Sequence is convergent, }\mbox{limit } =0 \end{align} or \begin{align} & u_{n}=\lim_{n \to \infty}\frac{n}{\left(n+1\right)\left(n+2\right)} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\frac{n}{n^{2}+3n+2} \\ & \qquad \quad \; = 0\\ & \therefore \mbox{Sequence is convergent, }\mbox{limit } =0 \end{align}

6) \frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{4}},\frac{1}{\sqrt{5}},\ldots u_{n}=\frac{1}{\sqrt{n+1}},\mbox{convergent}, \mbox{limit} =0

  • \begin{align} & u_{n}=\frac{1}{\sqrt{n+1}} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\frac{1}{\sqrt{n+1}} \\ & \qquad \quad \; = 0\\ & \therefore \mbox{Sequence is convergent, }\mbox{limit } =0 \end{align}

7) 1+\frac{1}{2},1+\frac{1}{4},1+\frac{1}{8},\ldots u_{n}=1+\frac{1}{2n},\mbox{convergent}, \mbox{limit} =1

  • \begin{align} & u_{n}=1+\frac{1}{2n} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}1+\cancelto{0}{\frac{1}{2n}} \\ & \qquad \quad \; = 1\\ & \therefore \mbox{Sequence is convergent, }\mbox{limit } =1 \end{align}

8) 1+\frac{1}{1},2+\frac{1}{2},3+\frac{1}{3},\ldots u_{n}=n+\frac{1}{n},\mbox{divergent}

  • \begin{align} & u_{n}=n+\frac{1}{n} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}n+\cancelto{0}{\frac{1}{n}} \\ & \qquad \qquad \; \mbox{does not exist}\\ & \therefore \mbox{Sequence is divergent} \end{align}

9) 9,16,25,36,\ldots u_{n}=\left(n+2\right)^{2},\mbox{divergent}

  • \begin{align} & u_{n}=\left(n+2\right)^{2} \\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\left(n+2\right)^{2} \\ & \qquad \qquad \; \mbox{does not exist}\\ & \therefore \mbox{Sequence is divergent} \end{align}

10) 4,1,-2,-5,\ldots u_{n}=7-3n,\mbox{divergent}\,

  • \begin{align} &\underbrace{4,1,-2,-5}_{\mbox{AP},a=4,r=-3} \\ & u_{n} = 4 +\left(n-1\right)\left(-3\right)\\ &\quad = 7-3n\\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}7-3n \\ & \qquad \qquad \; \mbox{does not exist}\\ & \therefore \mbox{Sequence is divergent} \end{align}

11) 1,-2,4,-8,\ldots u_{n}=\left(-2\right)^{n-1},\mbox{divergent}\,

  • \begin{align} &1,-2,4,-8,\ldots=\left(-2\right)^{0},\left(-2\right)^{1},\left(-2\right)^{2},\left(-2\right)^{3},\ldots \\ &u_{n}\quad = \left(-2\right)^{n-1}\\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\left(-2\right)^{n-1} \\ & \qquad \qquad \; \mbox{does not exist}\\ & \therefore \mbox{Sequence is divergent} \end{align}

12) 1,\frac{3}{2},\frac{9}{4},\frac{27}{8},\ldots u_{n}=\left(\frac{3}{2}\right)^{n-1},\mbox{divergent}\,

  • \begin{align} &1,\frac{3}{2},\frac{9}{4},\frac{27}{8},\ldots=\left(\frac{3}{2}\right)^{0},\left(\frac{3}{2}\right)^{1},\left(\frac{3}{2}\right)^{2},\left(\frac{3}{2}\right)^{3},\ldots \\ &u_{n}\quad = \left(\frac{3}{2}\right)^{n-1}\\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\left(\frac{3}{2}\right)^{n-1} \\ & \qquad \qquad \; \mbox{does not exist}\\ & \therefore \mbox{Sequence is divergent} \end{align}

13) \frac{2}{3},\frac{4}{9},\frac{8}{27},\ldots u_{n}=\left(\frac{2}{3}\right)^{n},\mbox{convergent}, \mbox{limit} =0

  • \begin{align} &\frac{2}{3},\frac{4}{9},\frac{8}{27},\ldots=\left(\frac{2}{3}\right)^{1},\left(\frac{2}{3}\right)^{2},\left(\frac{2}{3}\right)^{3},\ldots \\ &u_{n}\quad = \left(\frac{2}{3}\right)^{n}\\ & \lim_{n \to \infty} u_{n} = \lim_{n \to \infty}\left(\frac{2}{3}\right)^{n} \\ & \qquad \qquad \;=0\\ & \therefore \mbox{Sequence is convergent, }\mbox{limit } =0 \end{align}
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