Evaluate the sum ​

As n grows toward infinity, the 1/n+1 vanishes to 0, so the sum of the infinite series is 3/2

sumfor(n, 1, 1000000, 6 / ((2*n + 1)^2 - 1)) = 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

$S\quad{=\quad}\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{(2n+1)^2-1}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{4n^2+4n}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{4n(n+1)}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac32\bigg(\dfrac{1}{n(n+1)}\bigg)\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac32\bigg(\dfrac{1}{n}-\dfrac{1}{n+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}- \dfrac{1}{4}+\dots+\dfrac{1}{m-1}-\dfrac{1}{m}+\dfrac{1}{m}-\dfrac{1}{m+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(\dfrac11-\dfrac1{m+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(1-\dfrac1{m+1}\bigg)\\~\\~\\ S\quad=\quad\lim\limits_{m\rightarrow\infty}S_m\\~\\~\\ S\quad=\quad\lim\limits_{m\rightarrow\infty}\dfrac32\bigg(1-\dfrac1{m+1}\bigg)\\~\\~\\ S\quad=\quad\dfrac32\bigg(1-0\bigg)\\~\\~\\ S\quad=\quad\dfrac32$_