Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k^2+k+1}=\frac{\sqrt{3}\pi}{3}\cdot\tanh\left(\frac{\sqrt{3}\pi}{2}\right).$$

share|cite|improve this question
+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
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
up vote 16 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,


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 $A+C=0\implies A=-C=1$


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

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

If we set $U(m)=\dfrac{2m-1}{m^2-m+1},U(m+1)=\dfrac{2(m+1)-1}{(m+1)^2-(m+1)+1}=\dfrac{2m+1}{m^2+m+1}$

$$\implies T(n)=U(n)-U(n+1)$$

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

$$\implies2\sum_{k=1}^{\infty} \frac{k^2-1}{k^4+k^2+1}=U(1)=\cdots$$

share|cite|improve this answer
Could anybody please verify this? – lab bhattacharjee Feb 15 '13 at 15:01
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
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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.