# How to find sum of infinite series of non-geometric series?

I am familiar with radius of convergence, power series, Taylor series/Maclaurin series, and fundamental infinite series convergence tests (ratio, root, integral, comparison, etc.) introduced in a first semester college Calculus class. I have always had to determine convergence or divergence for most non-geometric or arithmetic series but I do not know a process to find $$\sum\limits_{k=1}^\infty \frac{(-1)^{k+1}k^2}{k^3+1}.$$ $\DeclareMathOperator{\sech}{sech}$ I have put the answer through Wolfram|Alpha and got $$\frac{1}{3}\left ( 1-\ln(2)+\pi \sech\left ( \frac{\sqrt3}{2}\pi \right ) \right ).$$

• $\checkmark \surd$ – four_lines Aug 6 '15 at 22:09
• Why do you want to compute the exact value of this series? DIfferent series need different approaches, and often it is not possible to find a closed form... – Crostul Aug 6 '15 at 22:15
• As a further comment, often series like this are related to other series like sum of inverse squares or the series for natural log, and the relationships are found by brute force computer work using identities, or complicated hypergeometric expressions. There's no telling how long such a proof could take for a particular formula. Do you really want to mess with this? It will probably be challenging for any response to come up with an "easy" way to evaluate the limit, and even if they did, it probably wouldn't help you evaluate other messed up series that Wolphram can magically evaluate. – user2566092 Aug 6 '15 at 22:20
• I only want to understand the approach that led to the answer. I start school next week and needed some refresher calculus for my DSP class. – Drew Aug 6 '15 at 22:20
• That specific one? Perhaps try partial fractions, and replace $\frac1{k+\alpha}$ with $\int_0^1 x^{k+\alpha-1}dx$… I feel like this'll get you a few geometric series. – Akiva Weinberger Aug 6 '15 at 22:21

$$\newcommand{\sech}{\operatorname{sech}}$$If we set $$\alpha=\frac{-1+i\sqrt3}2$$, we get $$\frac{1/3}{k+1}+\frac{1/3}{k+\alpha}+\frac{1/3}{k+\overline\alpha}=\frac{k^2}{k^3+1}\tag{0}$$ Therefore, \begin{align} \sum_{k=1}^\infty(-1)^{k+1}\frac{k^2}{k^3+1} &=\frac13\sum_{k=1}^\infty(-1)^{k+1}\left(\color{#C00000}{\frac1{k+1}}+\color{#00A000}{\frac1{k+\alpha}+\frac1{k+\overline\alpha}}\right)\tag{1}\\ &=\frac13\left(\color{#C00000}{1-\log(2)}+\color{#00A000}{\sum_{k\in\mathbb{Z}}\frac{(-1)^{k+1}}{k-\frac12+i\frac{\sqrt3}2}}\right)\tag{2}\\ &=\frac13\left(1-\log(2)+\pi\csc\!\left(\pi\left(\frac12-i\frac{\sqrt3}2\right)\right)\right)\tag{3}\\[6pt] &=\frac13\left(1-\log(2)+\pi\sech\!\left(\pi\frac{\sqrt3}2\right)\right)\tag{4} \end{align} Explanation:
$$(1)$$: use the partial fractions from $$(0)$$
$$(2)$$: sum of the alternating harmonic series is $$\log(2)$$
$$\hphantom{\text{(2):}}$$ and rewrite two unidirectional sums as a bidirectional sum
$$(3)$$: use $$(6)$$ from this answer
$$(4)$$: $$\sec\left(\pi i\frac{\sqrt3}2\right)=\sech\left(\pi\frac{\sqrt3}2\right)$$

The first step is to decompose $\frac{k^2}{k^3+1}$ as $\frac13\frac{1}{k+1}+\frac13\frac{2k-1}{k^2-k+1}$.

The first sum is easily evaluated as $$\frac13\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1}=\frac13(1-\ln2).$$

The second one, $$\frac13\sum_{k=1}^{\infty} \frac{(-1)^{k+1}(2k-1)}{k^2-k+1}=\frac13\pi\operatorname{sech}\frac{\sqrt{3}\pi}{2}$$

was already done on this site.
It follows from taking the logarithmic derivative of the Weierstrass product form of cosine : $$\cos(\pi x)=\prod_{k=0}^{\infty}\left( 1-\frac{4x^2}{(2k+1)^2}\right)$$