Evaluate the sum $\sum_{k=1}^\infty \big(\frac{1}{36k^2-1}+\frac{2}{(36k^2-1)^2}\big)$

I have to evaluate the series: $$\sum_{k=1}^\infty \left(\frac{1}{36k^2-1}+\frac{2}{(36k^2-1)^2}\right).$$

I tried using the identity $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$ but I got stuck.

• Did you try partial fractions? – abiessu Mar 2 '16 at 16:08
• Hint: $36k^2-1=(6k+1)(6k-1)$ – Nikunj Mar 2 '16 at 16:08
• @Nikunj The sequence in the problem converges to an irrational number, notably $\frac{\pi^2-9}{18}$. So partial fractions are unlikely to help very much, I think. – S.C.B. Mar 2 '16 at 16:10
• $\frac{1}{36k^2-1}=\frac{1}{2}\frac{(6k+1)-(6k-1)}{(6k+1)(6k-1)}$ – Nikunj Mar 2 '16 at 16:12
• Write $$1=\dfrac{6k+1-(6k-1)}2$$ and $$2=\dfrac{\{6k+1-(6k-1)\}^2}2$$ – lab bhattacharjee Mar 2 '16 at 16:24

1 Answer

A more general approach, proving the identity: $$\sum_{k\geq 1}\frac{k^2+x^2}{(k^2-x^2)^2}=-\frac{1}{2x^2}+\frac{\pi^2}{2\sin^2(\pi x)}\tag{1}$$ for any $x\in(-1,1)$ is to check that both the LHS and the RHS of $(1)$ are meromorphic functions with the same-behaving singularities at any $x\in\mathbb{Z}\setminus\{0\}$. For instance, $(1)$ is just a consequence of: $$\sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^2}=\frac{\pi^2}{\sin^2(\pi z)}\tag{2}$$ since $\frac{1}{(k-x)^2}+\frac{1}{(k+x)^2} = 2\cdot\frac{k^2+x^2}{(k^2-x^2)^2}$. $(2)$ can be seen as a consequence of the reflection formula for the $\Gamma$ function, too, since: $$\sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^2}= \psi'(-z)+\psi'(1+z). \tag{3}$$ If we plug $x=\frac{1}{6}$ in $(1)$, we get:

$$\sum_{k\geq 1}\frac{36k^2+1}{(36k^2-1)^2} = \color{red}{\frac{\pi^2-9}{18}}\tag{4}$$

and it is possible to derive such identity also through Fourier-analytic methods.

Still another way is to notice that our series is just $\frac{1}{2}\cdot L(\chi,2)$ where $\chi$ is the principal Dirichlet character $\!\!\pmod{6}$, hence our series depends on $\text{Li}_2$ evaluated at the sixth roots of unity.

An elementary proof of $(4)$ comes from the inclusion-exclusion principle: $$\sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}+\frac{1}{(6k+5)^2}\right) = \sum_{\substack{n\geq 1 \\ 2\nmid n \\ 3\nmid n}}\frac{1}{n^2} = \sum_{n\geq 1}\frac{1}{n^2}-\sum_{\substack{n\geq 1 \\ 2\mid n}}\frac{1}{n^2}-\sum_{\substack{n\geq 1 \\ 3\mid n}}\frac{1}{n^2}+\sum_{\substack{n\geq 1 \\ 6\mid n}}\frac{1}{n^2}$$ gives:

$$\sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}+\frac{1}{(6k+5)^2}\right) = \left(1-\frac{1}{4}-\frac{1}{9}+\frac{1}{36}\right) = \frac{2}{3}\cdot\zeta(2) = \color{red}{\frac{\pi^2}{9}}. \tag{5}$$