# How to compute $\sum_{k = 1}^{\infty} \frac{1}{k^2-\frac{1}{16}}$? [duplicate]

How do I compute $$\sum_{k = 1}^{\infty} \frac{1}{k^2-\frac{1}{16}}$$ ?

Mathematica says the sum converges and it somewhat looks like the Basel problem, but so far I do not know how to approach it.

A common way to solve the usual Basel problem is given here (proof 1).

By some clever tricks using Euler's formula which you can check out at the link above, it follows that

\begin{align*} \frac{1}{1+x^2}+\frac{1}{4+x^2}+\frac{1}{9+x^2}+\cdots&=-\frac{1}{2x^2}+\frac{\pi\cos(-i\pi x)}{2ix\sin(-i\pi x)} \end{align*} and at this point we usually plug in $$x=0$$ to get (in a limiting way) the value $$\pi^2/6$$.

In your case, we can put $$x=\frac i4$$ to get that the sum is $$\boxed{8-2\pi}$$.

Here's an elementary derivation that uses partial fractions, telescoping, and the $$\arctan$$ series: \begin{align} \sum_{k=0}^\infty \frac{1}{k^2-1/16} &=\sum_{k=0}^\infty \frac{16}{(4k+1)(4k-1)}\\ &=\sum_{k=0}^\infty \frac{16}{(2k+1)(2k-1)}\cdot\frac{1+(-1)^k}{2}\\ &=8\sum_{k=0}^\infty \frac{1}{(2k+1)(2k-1)}+8\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)(2k-1)}\\ &=4\sum_{k=0}^\infty \left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)+4\sum_{k=0}^\infty \frac{(-1)^k}{2k-1}-4\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\\ &=-4+4(-1-\pi/4)-4\pi/4\\ &=-8-2\pi, \end{align} so $$\sum_{k=1}^\infty \frac{1}{k^2-1/16} = -8-2\pi -\frac{1}{0^2-1/16}=8-2\pi.$$

You may use the Residue Theorem for Infinite Series to evaluate this sum: http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf

$$S=\sum_{k=1}^{\infty}\frac{1}{k^2-\frac{1}{16}}=8+\frac{1}{2}\sum_{k=-\infty}^{\infty}\frac{1}{k^2-\frac{1}{16}}$$

$$S=8+\frac{1}{2}\left[\lim_{z\to\frac{1}{4}}\frac{-\pi \cot(\pi z)(z-\frac{1}{4})}{z^2-\frac{1}{16}}+\lim_{z\to-\frac{1}{4}}\frac{-\pi \cot(\pi z)(z+\frac{1}{4})}{z^2-\frac{1}{16}}\right]$$

$$S=8-\frac{\pi}{2}\left[2\cot\left(\frac{\pi}{4}\right)-2\cot\left(-\frac{\pi}{4}\right)\right]=2\left(4-\pi\right)$$

The partial sums are also interesting since $$S_n=\sum_{k = 1}^{n} \frac{1}{k^2-\frac{1}{16}}=-2 \left(-\psi \left(n+\frac{3}{4}\right)+\psi \left(n+\frac{5}{4}\right)-\psi \left(\frac{5}{4}\right)+\psi \left(\frac{3}{4}\right)\right)$$ whic reduce to $$S_n=8-2\pi+2 H_{n-\frac{1}{4}}-2 H_{n+\frac{1}{4}}=8-2\pi-\frac{1}{n}+\frac{1}{2 n^2}-\frac{3}{16 n^3}+\frac{1}{32 n^4}+O\left(\frac{1}{n^5}\right)$$

Use it for $$n=5$$ $$S_5=\frac{22338896}{14549535}\approx 1.5353684$$ while the above truncated series gives $$S_5 \sim \frac{156371}{20000}-2 \pi\approx 1.5353647$$