[Math] $\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}$ and $ S_i =\sum _{n=1}^{\infty} \frac{i} {(36n^2-1)^i}$ . Find $S_1 + S_2 $

Question

I know to find sum of series using method of difference. I tried sum of write the term as (6n-1)(6n+1). i don't know how to proceed further.

Answer 1

Expand and split $$ \begin{align} \sum_{n=1}^{\infty}\frac{1}{n^2}&=\left(\color{blue}{\frac{1}{1^2}}+\color{red}{\frac{1}{2^2}}\right)+\left(\color{blue}{\frac{1}{3^2}}+\color{red}{\frac{1}{4^2}}\right)+\left(\color{blue}{\frac{1}{5^2}}+\color{red}{\frac{1}{6^2}}\right)+\cdots \\[2mm] &=\color{blue}{\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}}+\color{red}{\sum_{n=1}^{\infty}\frac{1}{\left(2n\right)^2}} \\[2mm] &=\color{blue}{\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}}+\color{red}{\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}} \end{align} \\ \space\Rightarrow\space \boxed{\color{blue}{\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}}=\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{8}} $$
Expand again $$ \begin{align} \sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}&=1+\left(\color{blue}{\frac{1}{{3}^2}}+\color{red}{\frac{1}{{5}^2}}+\color{magenta}{\frac{1}{{7}^2}}\right)+\left(\color{blue}{\frac{1}{{9}^2}}+\color{red}{\frac{1}{{11}^2}}+\color{magenta}{\frac{1}{{13}^2}}\right)+\left(\color{blue}{\frac{1}{{15}^2}}+\color{red}{\frac{1}{{17}^2}}+\color{magenta}{\frac{1}{{19}^2}}\right)+\cdots \\[2mm] &=1+\color{blue}{\sum_{n=1}^{\infty}\frac{1}{\left(6n-3\right)^2}}+\color{red}{\sum_{n=1}^{\infty}\frac{1}{\left(6n-1\right)^2}}+\color{magenta}{\sum_{n=1}^{\infty}\frac{1}{\left(6n+1\right)^2}} \\[2mm] &=1+\color{blue}{\frac{1}{9}\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}}+\color{red}{\sum_{n=1}^{\infty}\frac{1}{\left(6n-1\right)^2}}+\color{magenta}{\sum_{n=1}^{\infty}\frac{1}{\left(6n+1\right)^2}} \end{align} \\ \space\Rightarrow\space \boxed{\color{red}{\sum_{n=1}^{\infty}\frac{1}{\left(6n-1\right)^2}}+\color{magenta}{\sum_{n=1}^{\infty}\frac{1}{\left(6n+1\right)^2}}=\frac{8}{9}\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}-1=\frac{\pi^2}{9}-1} $$
Simplify $$ \begin{align} \frac{2}{\left(36n^2-1\right)^2}&=\frac{1}{2}\left[\frac{2}{(6n-1)(6n+1)}\right]^2=\frac{1}{2}\left[\frac{1}{6n-1}-\frac{1}{6n+1}\right]^2 \\[2mm] &=\frac{1/2}{\left(6n-1\right)^2}+\frac{1/2}{\left(6n+1\right)^2}-\frac{1}{36n^2-1} \end{align} $$ Claculate $$ \begin{align} \color{red}{S_1+S_2}&=\sum_{n=1}^{\infty}\frac{1}{36n^2-1}+\sum_{n=1}^{\infty}\frac{2}{\left(36n^2-1\right)^2}=\sum_{n=1}^{\infty}\frac{1/2}{\left(6n-1\right)^2}+\sum_{n=1}^{\infty}\frac{1/2}{\left(6n+1\right)^2} \\[2mm] &=\frac{1}{2}\sum_{n=1}^{\infty}\left[\frac{1}{\left(6n-1\right)^2}+\frac{1}{\left(6n+1\right)^2}\right]=\color{red}{\frac{\pi^2}{18}-\frac{1}{2}} \end{align} $$