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How would you go about evaluating:$$\sum_{n=2}^\infty\frac{(-1)^n}{n^2-n}$$

I split it up to $$\sum_{n=2}^\infty\left[(-1)^n\left(\frac{1}{n-1}-\frac{1}{n}\right)\right]$$

but I'm not sure what to do from here. If the $(-1)^n$ term wasn't there then it would be a simple telescoping series but the alternating bit causes trouble.

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  • $\begingroup$ Do you know the alternating series test? $\endgroup$ – Tyr Curtis Mar 19 '15 at 17:52
  • $\begingroup$ I do. The problem is not convergence, I've already verified that, it's evaluating the actual sum. $\endgroup$ – dd19 Mar 19 '15 at 17:56
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    $\begingroup$ Oh, sorry my mistake! Try the Taylor expansion of the natural logarithm. Well, Taylor series of $x\mapsto \ln(1+x)$. $\endgroup$ – Tyr Curtis Mar 19 '15 at 18:05
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Hint

We have

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\ln2$$ and we can prove it by several ways.

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  • $\begingroup$ So can I just break $\sum_{n=2}^\infty\frac{(-1)^n}{n-1}+\frac{(-1)^{n+1}}{n}$ into $\sum_{n=2}^\infty\frac{(-1)^n}{n-1}+\sum_{n=2}^\infty\frac{(-1)^{n+1}}{n}$ $\endgroup$ – dd19 Mar 19 '15 at 22:54
  • $\begingroup$ Yes because the two series are convergent. $\endgroup$ – user63181 Mar 19 '15 at 22:56
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\begin{align} & \sum_{n=2}^\infty (-1)^n\left(\frac{1}{n-1}-\frac{1}{n}\right) \\[10pt] = {} & \underbrace{\left(1-\frac 1 2 \right) - \left( \frac 1 2\right.}_{=\,0} - \left.\frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) - \left( \frac 1 4 - \frac 1 5\right) + \left( \frac 1 5 - \frac 1 6 \right) - \cdots \\[10pt] = {} & \frac 2 3 - \frac 2 4 + \frac 2 5 - \frac 2 6 + \cdots \\[10pt] = {} & 2\int_0^1 \left( u^2 - u^3 + u^4 - u^5 + \cdots \right) \,du = 2\int_0^1 \frac{u^2}{1+u} \, du \\[10pt] = {} & 2 \int_0^1 \left( u - 1 + \frac 1 {u+1} \right)\,du = \text{etc.} \end{align}

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  • $\begingroup$ $-(-\frac13) + \frac13 \neq 0$ $\endgroup$ – jameselmore Mar 19 '15 at 18:03
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$$\sum_{n=2}^{+\infty}\frac{(-1)^n}{n^2-n}=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(n+1)(n+2)}$$ and if we define $A_N$ as: $$ A_N = \sum_{n=0}^{N}\frac{(-1)^n}{n+1}$$ we have: $$\sum_{n=0}^{N}\frac{(-1)^n}{(n+1)(n+2)}=\sum_{n=0}^{N}(-1)^n\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=A_N+A_{N+1}-1,$$ but since $A_N\to \log 2$ as $N\to +\infty$, it follows that: $$\sum_{n=2}^{+\infty}\frac{(-1)^n}{n^2-n}=\color{red}{-1+\log 4}.$$

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