How to evaluate $\sum_{n=1}^{\infty} \frac{1}{n^2+4n+1}$

Question

How would we show that $$\sum_{n=1}^{\infty} \frac{1}{n^2+4n+1} = \frac{1}{6}(-2 -\pi \sqrt3 \cot(\pi \sqrt3))$$

I'm not sure how I'd go about it, but got the answer off wolfram alpha. Could someone lead me in the right direction?-series

Answer 1

The presence of the cotangent function together with the aspect of the general term of the given series suggest relations with the eulerian development of the cotangent:

for any complex number $z \in \mathbb{C}\setminus \pi\mathbb{Z}$ one has the series development: $$\mathrm{ctg}z=\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2-\pi^2n^2}.$$

To begin with, since $n^2+4n+1=(n+2)^2-3$ the given series can be rewritten as: $$\sum_{n=1}^{\infty}\frac{1}{n^2+4n+1}=\sum_{n=3}^{\infty}\frac{1}{n^2-3},$$ through a very clear bijective change of summation variable. Let us denote the sum of the series in question by $s$.

Taking $z=\pi\sqrt{3}$ in the forementioned development we obtain: $$\mathrm{ctg}(\pi\sqrt{3})=\frac{1}{\pi\sqrt{3}}+\frac{2\sqrt{3}}{\pi}\sum_{n=1}^{\infty}\frac{1}{3-n^2}=\frac{1}{\pi\sqrt{3}}+\frac{2\sqrt{3}}{\pi}\left(\frac{1}{2}+\frac{1}{-1}-s\right)$$ whence after an elementary rearrangement we gather: $$\sum_{n=1}^{\infty}\frac{1}{n^2+4n+1}=-\frac{1}{6}\left(2+\pi\sqrt{3}\mathrm{ctg}\left(\pi\sqrt{3}\right)\right).$$

Answer 2

$$S=\sum_{1}^{\infty} \frac{1}{n^2+4n+1}= \frac{1}{b-a} \left(\sum_{n=0}^{\infty} \frac{1}{n+a}-\frac{1}{n+b}\right) dx.$$ Here $a=2-\sqrt{3}, b=2+\sqrt{3}.$ Next, using integral tepresentation of: $1/z=\int_{0}^{1}t^{z-1} dt$, we get $$S=(b-a)^{-1}\int_{0}^{1}\sum_{n=1}^{\infty} [t^{n+a-1}-t^{n+b-1}] dt = (b-a)^{-1}\int_{0}^{1} \frac{t^{a}-t^{b}}{1-t}dt.$$ We have used $\sum_{1}^{\infty} z^n=\frac{z}{1-z},~if~|z|<1.$ Next using the definition of Harmonic numbers we can write $$S=H_b-H_a$$ I may get back again. Can some one take it from here?