How to show that $$\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$$ ?

My try:

We have $$n+3n+1=\left(n+\frac{3+\sqrt{5}}{2}\right)\left(n+\frac{3-\sqrt{5}}{2}\right),$$ so $$\frac{1}{n^2+3n+1}=\frac{2}{\sqrt{5}}\left(\frac{1}{2n+3-\sqrt{5}}-\frac{1}{2n+3+\sqrt{5}}\right).$$ Then, I don't know how to proceed.

  • $\begingroup$ What makes you the greatest difficulty in this task? Which way went your transformations? $\endgroup$ – Tacet Dec 5 '14 at 23:51
  • $\begingroup$ This answer should be useful: math.stackexchange.com/questions/112161 . $\endgroup$ – anomaly Dec 6 '14 at 0:11
  • $\begingroup$ Duplicate of this. $\endgroup$ – Lucian Dec 6 '14 at 3:11

I think we can make some use of the residue theorem. Write $n^2+3 n+1 = (n+3/2)^2-5/4$ and the sum is

$$\sum_{n=1}^{\infty} \frac1{\left (n+\frac{3}{2} \right )^2-\frac{5}{4}} = \frac12 \sum_{n=-\infty}^{\infty} \frac1{\left (n+\frac{3}{2} \right )^2-\frac{5}{4}} - 1 +1$$

(To get the doubly infinite sum, I had to add back the $n=0$ and $n=-1$ terms, which happen to sum to zero.)

The sum on the RHS may be attacked via the residue theorem, using the following:

$$\sum_{n=-\infty}^{\infty} f(n) = - \pi \sum_k \operatorname*{Res}_{z=z_k} [f(z)\cot{\pi z} ] $$

where $z_k$ is a non-integer pole of $f$. The poles are at

$$z_{\pm} = -\frac{3}{2} \pm \frac{\sqrt{5}}{2} $$

The sum is then equal to

$$\frac{\pi}{2 \sqrt{5}} \left [\cot{\left (\frac{3 \pi}{2} - \frac{\sqrt{5} \pi}{2} \right )} - \cot{\left (\frac{3 \pi}{2} + \frac{\sqrt{5} \pi}{2} \right )} \right ] = \frac{\pi}{\sqrt{5}}\tan{\frac{\sqrt{5}\pi}{2}}$$


Note $$ n^2+3n+1=(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2 $$ and and hence \begin{eqnarray} \sum_{n=0}^\infty\frac{1}{n^2+3n+1}&=&\sum_{n=0}^\infty\frac1{(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2}\\ &=&\frac12\sum_{n=-\infty}^\infty\frac1{(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2}. \end{eqnarray} Then using the result from this, you will get the answer.


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