Evaluate the infinite summation

Question

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

I tried to factor the denominator and then using partial fraction $$\frac{n^2}{1+n^3}=\frac{n^2}{(n+1)(n^2-n+1)}$$ $$=\frac{2n-1}{3(n^2-n+1)}+\frac{1}{3(n+1)}$$ So our question now becomes $$\frac13\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2n-1}{(n^2-n+1)}+\frac{1}{(n+1)}$$ This is not a telescopic series. I wrote a few terms and have observed this. Now I'm stuck.

Any help is greatly appreciated.

Answer 1

Let $S$ be the sum with the quadratic in the denominator. As user Ron Gordon points out in the linked question,

$$\begin{align*} S &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} \\ &= -\frac12 \sum_{k=-\infty}^{\infty} \frac{(-1)^{k} (2 k-1)}{k^2-k+1} \end{align*}$$

Take out the $k=0$ term and split up the sum by the sign of the index, then condense into a single sum over positive indices.

$$\begin{align*} S &= \frac12 - \frac12 \left\{\sum_{k=-\infty}^{-1} + \sum_{k=1}^\infty\right\} (-1)^k \frac{2k-1}{k^2-k+1} \\ &= \frac12 - \frac12 \sum_{k=1}^\infty (-1)^{-k} \frac{-2k-1}{k^2+k+1} - \frac12 \sum_{k=1}^\infty (-1)^k \frac{2k-1}{k^2-k+1} \\ &= \frac12 - \frac12 \sum_{k=1}^\infty (-1)^k \frac{k^2-1}{k^4+k^2+1} \\ &= \frac12 + \frac12 \sum_{k=1}^\infty (-1)^{k+1} \frac{k^2-1}{k^4+k^2+1} \end{align*}$$

Expand the summand into partial fractions.

$$\frac{k^2-1}{k^4+k^2+1} = \frac{e^{i\pi/3}}{e^{-i\pi/3}+k^2} + \frac{e^{-i\pi/3}}{e^{i\pi/3}+k^2}$$

Noting that $e^{\pm i\pi/3}=\mp e^{\mp i2\pi/3}$ and $\dfrac12=\cos\left(\dfrac\pi3\right)=\dfrac{e^{i\pi/3}+e^{-i\pi/3}}2$, we can write

$$\begin{align*} S &= \frac{e^{i\pi/3}+e^{-i\pi/3}}2 + \frac12 \sum_{k=1}^\infty (-1)^{k+1} \left[\frac{e^{i\pi/3}}{e^{i2\pi/3}-k^2} + \frac{e^{-i\pi/3}}{e^{-i2\pi/3}-k^2}\right] \\ &= \left[\frac1{2e^{-i\pi/3}} + \frac12 \sum_{k=1}^\infty (-1)^{k+1} \frac{e^{i\pi/3}}{\left(e^{i\pi/3}\right)^2-k^2}\right] + \left[\frac1{2e^{i\pi/3}} + \frac12 \sum_{k=1}^\infty (-1)^{k+1} \frac{e^{-i\pi/3}}{\left(e^{-i\pi/3}\right)^2-k^2}\right] \end{align*}$$

Comparing to the partial fraction expansion for $\csc$, the result follows,

$$S = \frac\pi2 \left[\csc\left(\pi e^{-i\pi/3}\right) + \csc\left(\pi e^{i\pi/3}\right)\right]$$

which we can rewrite as

$$S = \pi \sec\left(i\frac{\pi\sqrt3}2\right) = \pi\operatorname{sech}\left(\frac{\pi\sqrt3}2\right)$$

A quicker solution would be to complete the square in the denominator and compare against the partial fraction expansion of $\operatorname{sech}$.