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How to find $$\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}$$

I try something like this:

$$\begin{align*}\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}=\sum_{k=1}^{\infty}\frac{k^2}{k^4+k^2+1}-\sum_{k=1}^{\infty}\frac{1}{k^4+k^2+1}.\end{align*}$$

Using fact that $$\sum_{k = 1}^{n}{\frac{1}{k^4+k^2+1}}=\frac{1}{2}\cdot\frac{n+1}{n^2+n+1}+\frac{1}{2}\cdot\sum_{k = 1}^{n-1}{\frac{1}{k^2+k+1}}$$ we find that $$\begin{align*}\sum_{k=1}^{\infty}\frac{1}{k^4+k^2+1} &=\frac{1}{2}\cdot\sum_{k=1}^{\infty}{\frac{1}{k^2+k+1}}\\ &=\frac{1}{6}\left(\sqrt{3}\pi \tanh{\left(\frac{\sqrt{3}\pi}{2}\right)}-1\right).\end{align*}$$

But I don't know how to find $\displaystyle\sum_{k=1}^{\infty}\frac{k^2}{k^4+k^2+1}.$


If someone want to know how to evaluate $\displaystyle\sum_{k=0}^{\infty}\frac{1}{k^2+k+1}$:

First, $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k^2+k+1}=\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}.$$ Now, using "well-know" formula $$\displaystyle\cos(\phi)=\prod_{k=0}^{\infty}{\left( 1-\frac{4\phi^2}{(2k+1)^2\pi^2}\right)}$$ we find that $$\displaystyle\log (\cos(\phi))=\sum_{k=0}^{\infty}{\log\left( 1-\frac{4\phi^2}{(2k+1)^2\pi^2}\right)}$$ and then we attack with $\dfrac{d}{d\phi}$ and find $$\displaystyle\tan(\phi)=\sum_{k=0}^{\infty}{\frac{8\phi}{(2k+1)^2\pi^2-4\phi^2}}.$$ Let $\phi=\pi\alpha\cdot i$, then we get $$\displaystyle\tan(\pi\alpha\cdot i)=i\cdot\tanh(\pi\alpha)=i\cdot\sum_{k=0}^{\infty}{\frac{8\pi\alpha}{(2k+1)^2\pi^2+4\pi^2\alpha^2}}=\frac{2\alpha i}{\pi}\cdot\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\alpha^2}}.$$ So, we find that $$\displaystyle\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\alpha^2}}=\frac{\pi}{2\alpha}\cdot\tanh(\pi\alpha).$$ Let $ \alpha=\dfrac{\sqrt{3}}{2}.$ We get $$\displaystyle\sum_{k=0}^{\infty}{\frac{1}{\left(k+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}=\frac{\sqrt{3}\pi}{3}\cdot\tanh\left(\frac{\sqrt{3}\pi}{2}\right)$$ or $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k^2+k+1}=\frac{\sqrt{3}\pi}{3}\cdot\tanh\left(\frac{\sqrt{3}\pi}{2}\right).$$

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3  
+1 Nice question: a question and (a partial) answer all wrapped up in one package! (re edit: I simply "highlighted" your question)... –  amWhy Feb 15 '13 at 14:42
    
@amWhy Thank you. Nice edit, I like it :) –  Cortizol Feb 15 '13 at 14:55
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Hmm... $\frac{1}{k^2+k+1} + \frac{1}{k^2-k+1} = \frac{2(k^2+1)}{k^4+k^2+1}$ and $\frac{1}{k^2-k+1} = \frac{1}{(k-1)^2 + (k-1)+1}$. –  achille hui Feb 15 '13 at 15:04

1 Answer 1

up vote 12 down vote accepted

Using Partial Fraction Decomposition $$\frac{k^2-1}{k^4+k^2+1}=\frac{Ak+B}{k^2-k+1}+\frac{Ck+D}{k^2+k+1}$$

So, $k^2-1=k^3(A+C)+k^2(A+B-C+D)+k(A+B+C-D)+B+D$

Comparing the coefficients of different powers of $k$ in the above identity,

$A+C=0$

From $A+B+C-D=0,B+D=0$ and $B+D=-1\implies B=-D=-\frac12$

From $A+B-C+D=1\implies A-C=2$ and $A+C=0\implies A=-C=1$

So, $$\frac{k^2-1}{k^4+k^2+1}=\frac{k-\frac12}{k^2-k+1}-\frac{k+\frac12}{k^2+k+1}$$

$$\frac{2(k^2-1)}{k^4+k^2+1}=\frac{2k-1}{k^2-k+1}-\frac{2k+1}{k^2+k+1}=T(k)\text{ say}$$

So, $$T(n)=\frac{2n-1}{n^2-n+1}-\frac{2n+1}{n^2+n+1}$$

So, $$T(n+1)=\frac{2(n+1)-1}{(n+1)^2-(n+1)+1}-\frac{2(n+1)+1}{(n+1)^2+(n+1)+1}$$ $$=\frac{2n+1}{n^2+n+1}-\frac{2n+3}{n^2+3n+3}$$

Clearly, the first part of any term except the first term is cancelled by the last part of the previous term.

For example,

$T(1)=\frac11-\frac33, T(2)=\frac33-\frac57,T(3)=\frac57-\frac7{13}$

So, $$2\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}=1$$

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Could anybody please verify this? –  lab bhattacharjee Feb 15 '13 at 15:01
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Verified by hand and by mathematica. –  muzzlator Feb 15 '13 at 15:07
    
@muzzlator, thanks for your feedback. –  lab bhattacharjee Feb 15 '13 at 15:08
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Note that $\displaystyle \sum_{k=1}^{\infty} \frac{2k - 1}{k^2 - k + 1} = \sum_{k=0}^{\infty} \frac{2(k+1) - 1}{(k+1)^2 - (k+1) + 1}$ –  J.H. Feb 15 '13 at 15:10
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@muzzlator, I just tried to see what happens after Partial Fraction Decomposition and then identified the telescoping nature. –  lab bhattacharjee Feb 15 '13 at 15:17

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