Evaluate the sum \(\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots.\)
As n grows toward infinity, the 1/n+1 vanishes to 0, so the sum of the infinite series is 3/2
6 [ 1 / [ 3^2 - 1 ] + 1 / [ 5^2 - 1] + 1/[7^2 - 1] + 1/ [ 9^2 - 1 ] +....... ] =
6 [ 1 / 8 + 1 / 24 + 1/48 + 1/80 + ..... ] =
6 [ 1 / [ 2*4] + 1/ [ 4*6] + 1/[ 6*8] + 1/ [8*10] + ....... ] =
6 [ (1/4) ] [ 1 / [1 * 2] + 1/ [2 *3] + 1/ [ 3*4] + 1 / [ 4*5] + .....]
Note that 1/ [ n * (n + 1) ] = 1 / n - 1/[ n + 1]
So we can write
(6/4) [ ( 1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .......] =
(3/2) [ 1 + ( 1/2 -1/2) + (1/3 - 1/3) + (1/4 - 1/4) + ........ + (-1/n) ] =
(3/2) [ 1 - 1/n ] as n ⇒ infinity, 1/n ⇒ 0
So we have
(3/2 ) [ 1 - 0] =
(3/2) (1) =
3/2
\(S\quad{=\quad}\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{(2n+1)^2-1}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{4n^2+4n}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{4n(n+1)}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac32\bigg(\dfrac{1}{n(n+1)}\bigg)\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac32\bigg(\dfrac{1}{n}-\dfrac{1}{n+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}- \dfrac{1}{4}+\dots+\dfrac{1}{m-1}-\dfrac{1}{m}+\dfrac{1}{m}-\dfrac{1}{m+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(\dfrac11-\dfrac1{m+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(1-\dfrac1{m+1}\bigg)\\~\\~\\ S\quad=\quad\lim\limits_{m\rightarrow\infty}S_m\\~\\~\\ S\quad=\quad\lim\limits_{m\rightarrow\infty}\dfrac32\bigg(1-\dfrac1{m+1}\bigg)\\~\\~\\ S\quad=\quad\dfrac32\bigg(1-0\bigg)\\~\\~\\ S\quad=\quad\dfrac32\)_
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