$\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}$ 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 $$ \begin{align} \color{red}{S_1+S_2}&=\sum_{n=1}^{\infty}\frac{1}{36n^2-1}+\sum_{n=1}^{\infty}\frac{2}{\left(36n^2-1\right)^2}=\sum_{n=1}^{\infty}\frac{1/2}{\left(6n-1\right)^2}+\sum_{n=1}^{\infty}\frac{1/2}{\left(6n+1\right)^2} \\[2mm] &=\frac{1}{2}\sum_{n=1}^{\infty}\left[\frac{1}{\left(6n-1\right)^2}+\frac{1}{\left(6n+1\right)^2}\right]=\color{red}{\frac{\pi^2}{18}-\frac{1}{2}} \end{align} $$

Answer 2

\begin{align*} S_1+S_2&=\sum_{n\geq 1}\frac{1}{36n^2-1}+\frac{2}{(36n^2-1)^2}\\ &=\sum_{n\geq 1}\frac{1}{2}\frac{1}{(6n-1)^2}+\frac{1}{2}\frac{1}{(6n+1)^2}\\ &=\frac{1}{2}\sum_{\substack{n\geq 5\\n\equiv \pm1\,\!\!\!\mod 6}}\frac{1}{n^2}\\ &=\frac{1}{2}\left[\sum_{n\geq 1}\frac{1}{n^2}-\sum_{\substack{n\geq 1\\n\equiv 0\!\!\!\mod 2}}\frac{1}{n^2}-\sum_{\substack{n\geq 1\\n\equiv 0\!\!\!\mod 3}}\frac{1}{n^2}+\sum_{\substack{n\geq 1\\n\equiv 0\!\!\!\mod 6}}\frac{1}{n^2}-1\right]\\ &=\frac{1}{2}\left[\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}-\frac{1}{9}\frac{\pi^2}{6}+\frac{1}{36}\frac{\pi^2}{6}-1\right]\\ &=\frac{\pi^2}{18}-\frac{1}{2}. \end{align*}

Answer 3

Assuming that you know about the polygamma functions, first write $$(36n^2-1)^2=(6n+1)^2(6n-1)^2$$ and perform a first partial fraction decomposition $$\frac 2{(36n^2-1)^2}=\frac{1}{2 (6 n+1)}-\frac{1}{2 (6 n-1)}+\frac{1}{2 (6 n+1)^2}+\frac{1}{2 (6 n-1)^2}$$ Computing the partial sum $$\Sigma_p=\sum_{n=1}^p\frac 2{(36n^2-1)^2}=$$ $$\Bigg[\frac{1}{12} \left(\psi ^{(0)}\left(p+\frac{7}{6}\right)-\psi ^{(0)}\left(\frac{7}{6}\right)\right) \Bigg]-\Bigg[\frac{1}{12} \left(\psi ^{(0)}\left(p+\frac{5}{6}\right)-\psi ^{(0)}\left(\frac{5}{6}\right)\right) \Bigg]+\Bigg[\frac{1}{72} \left(\psi ^{(1)}\left(\frac{7}{6}\right)-\psi ^{(1)}\left(p+\frac{7}{6}\right)\right) \Bigg]+\Bigg[\frac{1}{72} \left(\psi ^{(1)}\left(\frac{5}{6}\right)-\psi ^{(1)}\left(p+\frac{5}{6}\right)\right) \Bigg]$$

Now, using the asymptotics $$\Sigma_p=\frac{1}{72} \left(6 \psi ^{(0)}\left(\frac{5}{6}\right)-6 \psi ^{(0)}\left(\frac{7}{6}\right)+\psi ^{(1)}\left(\frac{5}{6}\right)+\psi ^{(1)}\left(\frac{7}{6}\right)\right)-\frac{1}{1944 p^3}+O\left(\frac{1}{p^4}\right)$$ Thus $$S_2=\frac{1}{72} \left(6 \psi ^{(0)}\left(\frac{5}{6}\right)-6 \psi ^{(0)}\left(\frac{7}{6}\right)+\psi ^{(1)}\left(\frac{5}{6}\right)+\psi ^{(1)}\left(\frac{7}{6}\right)\right)=-1+\frac{\pi }{4 \sqrt{3}}+\frac{\pi ^2}{18}$$

$$S_1+S_2=\frac{2 \pi ^2}{9}+\frac{\pi }{4 \sqrt{3}}-1\sim 1.64670 $$

Obviously, this is not the same result as in other answers.

I must say that I do not understand how, adding positive terms to $S_1$, we could have a vlaue smaller then $\frac{ \pi ^2}{6} \sim 1.64493$.