How to evaluate this sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{(n^2 + 3n + 1)(n^2 - 3n + 1)}$

Question

How to evaluate this sum $$\sum_{n=1}^{\infty} \frac{(-1)^n}{(n^2 + 3n + 1)(n^2 - 3n + 1)}$$

My attempt

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{(n^2 + 3n + 1)(n^2 - 3n + 1)}$$

$$= \sum_{n=1}^{\infty} \frac{(-1)^{n - 1 + 1}}{(n^2 + 2 \cdot \frac{3}{2}n + 1)(n^2 - 2 \cdot \frac{3}{2}n + 1)}$$

$$ = \sum_{n=1}^{\infty} \frac{(-1)^{n - 1} \cdot (-1)}{(n^2 + 2 \cdot \frac{3}{2}n + \frac{9}{4} - \frac{5}{4})(n^2 - 2 \cdot \frac{3}{2}n + \frac{9}{4} - \frac{5}{4})} $$

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

$$= - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{[(n + \frac{3}{2})^2 - (\frac{\sqrt{5}}{2})^2][(n - \frac{3}{2})^2 - (\frac{\sqrt{5}}{2})^2]}$$

$$ = -\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n + \frac{3}{2} - \frac{\sqrt{5}}{2})(n + \frac{3}{2} + \frac{\sqrt{5}}{2})(n - \frac{3}{2} - \frac{\sqrt{5}}{2})(n - \frac{3}{2} + \frac{\sqrt{5}}{2})}$$

$$ = - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{[n + \frac{3 - \sqrt{5}}{2}][n + \frac{3 + \sqrt{5}}{2}][n - \frac{3 + \sqrt{5}}{2}][n - \frac{3 - \sqrt{5}}{2}]}$$

$$= - \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{1}{[n + \frac{3 + \sqrt{5}}{2}][n - \frac{3 + \sqrt{5}}{2}]} \cdot \frac{1}{[n + \frac{3 - \sqrt{5}}{2}][n - \frac{3 - \sqrt{5}}{2}]}$$

$$= - \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{(3 + \sqrt{5} + n - n)}{(3 + \sqrt{5})[n + \frac{3 + \sqrt{5}}{2}][n - \frac{3 + \sqrt{5}}{2}]} \cdot \frac{(3 - \sqrt{5} + n - n)}{(3 - \sqrt{5})[n + \frac{3 - \sqrt{5}}{2}][n - \frac{3 - \sqrt{5}}{2}]}$$

$$= = -\frac{1}{(3+\sqrt{5})(3-\sqrt{5})} \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{\left(\frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} + n - n\right)}{\left[n + \frac{3 + \sqrt{5}}{2}\right]\left[n - \frac{3 + \sqrt{5}}{2}\right]} \cdot \frac{\left(\frac{3 - \sqrt{5}}{2} + \frac{3 + \sqrt{5}}{2} + n - n\right)}{\left[n + \frac{3 - \sqrt{5}}{2}\right]\left[n - \frac{3 - \sqrt{5}}{2}\right]}$$

$$= -\frac{1}{3^2 - (\sqrt{5})^2} \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{\left(n + \frac{3 + \sqrt{5}}{2} - n + \frac{3 + \sqrt{5}}{2}\right)}{\left[n + \frac{3 + \sqrt{5}}{2}\right]\left[n - \frac{3 + \sqrt{5}}{2}\right]} \cdot \frac{n + \frac{3 - \sqrt{5}}{2} - n + \frac{3 - \sqrt{5}}{2}}{\left[n + \frac{3 - \sqrt{5}}{2}\right]\left[n - \frac{3 - \sqrt{5}}{2}\right]}$$

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

Answer 1

Let's use a standard way of evaluation of such sums. Denoting $S$ the desired sum, we consider the integral in the complex plane along a big circle $C_R$ with the radius $R$ $$I_C=\oint_C\frac\pi{\sin\pi z}\frac{dz}{(z^2 + 3z + 1)(z^2 - 3z + 1)}\to 0 \,\,\text{at}\,\,R\to\infty$$ On the other hand, $$I_C=2\pi i\sum\operatorname{Res}\frac\pi{\sin\pi z}\frac1{(z^2 + 3z + 1)(z^2 - 3z + 1)}\,\,(\to 0)$$ We have simple poles inside our closed contour: at the points $z=0, \pm1, \pm2,...$ (the residues at these points give $\displaystyle \sum_{n=-\infty}^\infty\frac{(-1)^n}{(n^2 + 3n + 1)(n^2 - 3n + 1)}=2S+1\,\,$); and four poles at $z=\pm\frac32\pm\frac{\sqrt5}2$.

Evaluating the residues at these four poles, $$2S+1+\frac{2\pi}{\sin\pi(\frac32+\frac{\sqrt 5}2)}\frac1{3\sqrt5(3+\sqrt5)}+\frac{2\pi}{\sin\pi(-\frac32+\frac{\sqrt 5}2)}\frac1{3\sqrt5(3-\sqrt5)}=0$$ $$S=-\frac\pi{6\cos\frac{\pi\sqrt 5}2}-\frac12$$

Answer 2

A more elementary (high-school level) table-based proof that dispenses with complex analysis runs as follows.

Starting from: $$x^2+3x+1 =\left(x-\frac{1}{2} \left(-\sqrt{5}-3\right)\right) \left(x-\frac{1}{2} \left(\sqrt{5}-3\right)\right)$$ and $$x^2-3x+1 =\left(x+\frac{1}{2} \left(-\sqrt{5}-3\right)\right) \left(x+\frac{1}{2} \left(\sqrt{5}-3\right)\right)$$ so that we have

\begin{align} a(x)=\frac{1}{(x^2+3x+1)(x^2-3x+1)} &= \frac{3-x}{6 \left(x^2-3 x+1\right)}+\frac{x+3}{6 \left(x^2+3 x+1\right)} \\ &=\frac{1}{3\sqrt{5}}\left(\frac{x-3}{ -2 x+\sqrt{5}+3}+\frac{x-3}{2 x+\sqrt{5}-3}-\frac{x+3}{-2 x+\sqrt{5}-3}-\frac{x+3}{2 x+\sqrt{5}+3}\right) \\ &= \frac{1}{3\sqrt{5}}\left(\frac{\sqrt{5}-3}{2 \left(-2 x+\sqrt{5}+3\right)}-\frac{\sqrt{5}+3}{2 \left(2 x+\sqrt{5}-3\right)}-\frac{\sqrt{5}+3}{2 \left(-2 x+\sqrt{5}-3\right)}+\frac{\sqrt{5}-3}{2 \left(2 x+\sqrt{5}+3\right)}\right) \tag{1} \end{align}

It is a tabulated formula (Prudnikov et al., Integrals and Series, Vol I, eq. 5.1.25.6 p.685, 1998 edition) that:

$$\sum _{n=1}^{\infty } \frac{(-1)^n}{b^2\,n^2-a^2}=\frac{b-\pi a \csc \left(\frac{\pi a}{b}\right)}{2 a^2 b} \tag{2}$$

so that grouping together the first and last rational expression in $(1)$, then the second and the third and applying $(2)$ twice with $b=2$ and $a = \sqrt{5} \pm 3$, there finally comes, after some immediate algebra:

$$\sum_{n=1}^\infty (-1)^na(n)=-\frac{1}{6} \left(3+\pi \sec \left(\frac{\sqrt{5} \pi }{2}\right)\right)$$

Numerical verification (Mathematica 14):

NSum[(-1)^n / ((n^2 + 3 n + 1) (n^2 - 3 n + 1)), {n, 1, Infinity}]
0.0617817
N[1/6 (-3-\[Pi] Sec[(Sqrt[5] \[Pi])/2])]
0.0617817

Edit note

This method is quite general and works irrespective of coefficients $a, b, c$ for such "conjugate" sums of the type

$$\sum_{n=1}^\infty \frac{(-1)^n}{T(n)T(-n)},$$

where $T(x) = ax^2 + bx +x$.