Sequences Series Part2

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Summation Notation \sum

  • \sum^{n}_{r=1}u_{r} is read as
    • {\color{blue}\sum} sum
    • \sum{\color{blue}u_{r}} sum of u_{r}\,
    • \sum_{{\color{Blue}r=1}}u_{r} sum of u_{r}\, from r=1\,
    • \sum_{r=1}^{{\color{Blue}n}}u_{r} sum of u_{r}\, from r=1\, to r=n\, (note that r=\, is not written at the top part as it it unnecessary)

Now let's see how to write out the sum

  • \sum^{n}_{r=1}u_{r}
    • We can see from the above, that this means
      • the sum of all the u_{{\color{Blue}r}}\,'s, from
      • {\color{Blue}r}={\color{Red}1}\,, meaning from u_{{\color{Red}1}}\,
      • to {\color{Blue}r}={\color{Red}n}\,, meaning to u_{{\color{Red}n}}\,
    • In others words, sum from u_{1}\, to u_{n}\,
    • \sum^{n}_{r=1}u_{r} u_{1}+u_{2}+u_{3}+\ldots+ u_n{}\,

Notes

  • S_{n}\, have the same meaning, but is usually only used in AP & GP
  • In sequences, we used u_{n}\, to represent the general term, that is, any term in the sequence.
  • In series, however, u_{n}\, is reserved for the last term only. Thus, for the general term, we need to use another alphabet, thus we use u_{r}\,
  • General term represented by u_{r}\,
    • r=\, 1,2,3,\ldots, n\,

Examples

  • \sum^{5}_{r=1}u_{r} =u_{1}+u_{2}+u_{3}+u_{4}+u_{5}\,
    • \sum^{{\color{Blue}5}}_{r={\color{Red}1}}u_{r}=u_{{\color{Red}1}}+u_{2}+u_{3}+u_{4}+u_{{\color{Blue}5}}


  • \sum^{7}_{r=3}u_{r} =u_{3}+u_{4}+u_{5}+u_{6}+u_{7}\,
    • \sum^{{\color{Blue}7}}_{r={\color{Red}3}}u_{r}=u_{{\color{Red}3}}+u_{4}+u_{5}+u_{6}+u_{{\color{Blue}7}} You can see that this notation is more flexible than S_{n}\, in that it allows r\, to start from any integer and not just from 1\,


  • \sum^{2n}_{r=1}u_{r} =u_{1}+u_{2}+u_{3}+\ldots+u_{2n}
    • \sum^{{\color{Blue}2n}}_{r=1}u_{r}=u_{1}+u_{2}+u_{3}+\ldots+u_{{\color{Blue}2n}}


  • \sum^{2n}_{r=n+1}u_{r} =u_{n+1}+u_{2}+u_{3}+\ldots+u_{2n}\,
    • \sum^{{\color{Blue}2n}}_{r={\color{Red}n+1}}u_{r}=u_{{\color{Red}n+1}}+u_{2}+u_{3}+\ldots+u_{{\color{Blue}2n}}


Furthermore, this notation allows us to write exactly what is u_{r}\, right at the notation (compared to S_{n}\, where u_{n}\, have to be written seperately

Examples

  • \sum^{5}_{r=1}r =1+2+3+4+5\,
    • Note that \sum^{5}_{r={\color{Red}1}}r goes into \sum^{5}_{r=1}{\color{Red}r}, starting from 1\, to \sum^{{\color{Red}5}}_{r=1}r


  • \sum^{8}_{r=4}r^{2} =4^{2}+5^{2}+6^{2}+7^{2}+8^{2}\,
    • \sum^{{\color{Red}8}}_{r={\color{Red}4}}r^{2}={\color{Red}4}^{2}+{\color{Red}5}^{2}+{\color{Red}6}^{2}+{\color{Red}7}^{2}+{\color{Red}8}^{2}


  • \sum^{10}_{r=1}\left(r+1\right) =2+3+4+\ldots+11\,
    • It's best NOT to write \left(1+1\right)+\left(1+2\right)+\left(1+3\right) but instead count it directly, but carefully.


  • \sum^{20}_{r=1}\frac{r}{\left(r+1\right)\left(r+2\right)} =\frac{1}{2\cdot3}+\frac{2}{3\cdot4}+\frac{3}{4\cdot5}\ldots+\frac{20}{21\cdot22}


Writing Series using \sum

To be able to write a series using the notation \sum^{{\color{Red}n}}_{r=1}{\color{Blue}u_{r}}, we must

  • determine u_{r}\, Note that we usually choose an u_{r}\, so that it the series starts with r=1\,
  • determine n\,

Exercise 2

1) Write out the following sums

  • a) \sum_{r=1}^{5}r=\, 1+2+3+4+5\,


  • b) \sum_{r=1}^{6}\left(r+1\right)\left(r+2\right)=\, 2\cdot3+3\cdot4+4\cdot5+5\cdot6+6\cdot7+7\cdot8


  • c) \sum_{r=1}^{4}\frac{2r-1}{2r+1}=\, \frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9}\,


  • d) \sum_{r=5}^{10}r^{3}=\, 5^{3}+6^{3}+7^{3}+8^{3}+9^{3}+10^{3}\,


  • e) \sum_{r=1}^{6}\left(-1\right)^{r}r=\, -1+2-3+4-5+6\,


2) Express the following using \sum \, notation

  • a) 1+2+3+\ldots+100 \sum_{r=1}^{100} r
    • \begin{align}
& u_{r}=r\\
& u_{n}=100, \therefore n=100 \\
& \therefore 1+2+3+\ldots+100 = \sum_{r=1}^{100} r
\end{align}


  • b) 5+7+9+\ldots+99 \sum_{r=1}^{48} \left(2r+3\right)
    • \begin{align}
& u_{r}=2r+3\\
& u_{n}=99, \therefore n=48 \\
& \therefore 5+7+9+\ldots+99 = \sum_{r=1}^{48} \left(2r+3\right)
\end{align}


  • c) 1+4+9+\ldots+576 \sum_{r=1}^{24} r^{2}
    • \begin{align}
& u_{r}=r^{2}\\
& u_{n}=576, n^{2}=576 ,\therefore n=\sqrt{576}=24 \\
& \therefore 1+4+9+\ldots+576 = \sum_{r=1}^{24} r^{2}
\end{align}


  • d) 13+9+5+1+\ldots-27 \sum_{r=1}^{11} \left(17-4r\right)
    • \begin{align}
& u_{r}=17-4r\\
& u_{n}=-27, \therefore n=11 \\
& \therefore 13+9+5+1+\ldots-27 = \sum_{r=1}^{11} \left(17-4r\right)
\end{align}


  • e) 1+\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}+\frac{32}{243} \sum_{r=1}^{6} \left(\frac{2}{3}\right)^{r-1}
    • \begin{align}
& u_{r}=\left(\frac{2}{3}\right)^{r-1}, n=6 \\
& \therefore 1+\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}+\frac{32}{243} = \sum_{r=1}^{6} \left(\frac{2}{3}\right)^{r-1}
\end{align}


  • f) 3-9+27-81+243-729\, \sum_{r=1}^{6} -\left(-3^{r}\right)
    • \begin{align}
& u_{r}=3\left(-3\right)^{r-1}=-\left(-3^{r}\right), n=6\\
& \therefore 3-9+27-81+243-729 = \sum_{r=1}^{6} -\left(-3^{r}\right)
\end{align}


  • g) \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\ldots+\frac{100}{101}\, \sum_{r=1}^{100} \frac{r}{r+1}
    • \begin{align}
& u_{r}=\frac{r}{r+1}, n=100\\
& \therefore \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\ldots+\frac{100}{101} = \sum_{r=1}^{100} \frac{r}{r+1}
\end{align}


  • h) \frac{1}{2}+\frac{4}{5}+\frac{7}{8}+\frac{10}{11}+\ldots+\frac{151}{152}\, \sum_{r=1}^{51} \frac{3r-2}{3r-1}
    • \begin{align}
& u_{r}=\frac{3r-2}{3r-1}, n=51\\
& \therefore \frac{1}{2}+\frac{4}{5}+\frac{7}{8}+\frac{10}{11}+\ldots+\frac{151}{152} = \sum_{r=1}^{51} \frac{3r-2}{3r-1}
\end{align}


  • i) \frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{100^{2}}\, \sum_{r=1}^{97} \frac{1}{\left(r+2\right)^{2}}
    • \begin{align}
& u_{r}={\left(r+2\right)}^{2}\\
& n+3=100, n=97\\
& \therefore \frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{100^{2}} = \sum_{r=1}^{97} \frac{1}{\left(r+2\right)^{2}}
\end{align}


  • j) \sqrt{2}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\ldots+\sqrt{1600} \sum_{r=1}^{800}\sqrt{2n}
    • \begin{align}
& u_{r}=\sqrt{2r}\\
& 2n=1600, n=800\\
& \therefore \sqrt{2}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\ldots+\sqrt{1600} = \sum_{r=1}^{800}\sqrt{2n} 
\end{align}