5

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 $$\frac{1}{3}\left ( 1-\ln(2)+\pi \sech\left ( \frac{\sqrt3}{2}\pi \right ) \right ).$$

Clayton
  • 24,751
Drew
  • 380
  • $\checkmark \surd$ – four_lines Aug 06 '15 at 22:09
  • 1
    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 06 '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 06 '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 06 '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 06 '15 at 22:21
  • http://math.stackexchange.com/questions/576212/foxtrot-bill-amend-problems/576250#576250 – Ron Gordon Aug 07 '15 at 00:37
  • @RonGordon: I knew this seemed familiar... I did this problem when that Foxtrot was in the paper. I even wrote up a PDF with the answer. It was pretty close to my answer below. I was also playing with texture mapping in Mathematica and made this reproduction. – robjohn Aug 07 '15 at 00:44
  • @robjohn: it's really hard to keep track of what we answered, like, 2 years ago. Somehow, though, this one did stick out to me because of the Foxtrot angle. Bill Amend is the second coolest cartoonist out there. – Ron Gordon Aug 07 '15 at 00:49
  • @robjohn: very cool effect with the comic! – Ron Gordon Aug 07 '15 at 00:55

2 Answers2

5

$\newcommand{\sech}{\operatorname{sech}}$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
$(4)$: $\sec\left(\pi i\frac{\sqrt3}2\right)=\sech\left(\pi\frac{\sqrt3}2\right)$

Martin Sleziak
  • 53,687
robjohn
  • 345,667
3

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 : $$\cos(\pi x)=\prod_{k=0}^{\infty}\left( 1-\frac{4x^2}{(2k+1)^2}\right)$$

Community
  • 1
Noam Shalev - nospoon
  • 6,595