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I have to use

$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}=\log 2$

to compute

$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n(n+1)}$

Since,

$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n(n+1)} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n+1)}$

Now, $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} = \ -log 2 $

$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n+1)} = \log 2 -1 $

Hence , the answer seems to be $1 - 2 \log 2$.

  • 4
    Hint: $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$.2012-05-17

2 Answers 2

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Noting that

$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

we have

$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n(n+1)}= \sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n}-\frac{(-1)^{n}}{n+1}\right)$

The first part is obviously $-\log 2$ (from the definition above), so we have

$\sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n}-\frac{(-1)^{n}}{n+1}\right) =-(\log 2+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n+1})$

I'm sure you can take it from here (a substitution may help).

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    It's interesting to note that original series is absolutely convergent, but $\sum a_n$ and $\sum b_n$ are conditionally convergent. By the Riemann rearrangement theorem, they can be resummed to any real, but the LHS is of course invariant. So what's the deeper justification?2012-05-18
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HINT:

Consider partial fraction decomposition.