# ∑∞n=11n2=π26\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6} and Si=∑∞n=1i(36n2−1)i S_i =\sum _{n=1}^{\infty} \frac{i} {(36n^2-1)^i} . Find S1+S2S_1 + S_2

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.

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S1+S2=∑n≥1136n2−1+2(36n2−1)2=∑n≥1121(6n−1)2+121(6n+1)2=12∑n≥5n≡±1mod61n2=12[∑n≥11n2−∑n≥1n≡0mod21n2−∑n≥1n≡0mod31n2+∑n≥1n≡0mod61n2−1]=12[π26−14π26−19π26+136π26−1]=π218−12.\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*}

how can i explain 4th step, if one does not know modulo math.can it be explained in easier way
– raj
2 days ago