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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 $1 - 2 \log 2$.

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    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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    I am guessing the $-1$ was for the fact that you need justification for $\sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n}-\frac{(-1)^{n}}{n+1}\right) =-(\log 2+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n+1})$ and is a common mistake students make. (I didn't upvote, in case you are wondering).2012-05-17
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    @Aryabhata Which $-1$ are you referring to?2012-05-17
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    Someone downvoted this answer, and for a valid reason.2012-05-17
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    @Argon: Thanks for your reply. Can you please verify the solution.2012-05-17
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    @Aryabhata I imagine you are referring to how I found the sum to equal $\log$ instead of $-\log$, which I corrected. Is there something currently wrong with my answer?2012-05-17
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    @Argon: No. You are moving whole series around without justification(which is probably important pedantically speaking, as this is homework). Of course, what you did is easily justified. See my first comment.2012-05-17
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    Well, $$\log x = 1-\frac{1}{2}+\frac{1}{3}-\cdots=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$$ Now, $$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}=-1+\frac{1}{2}-\frac{1}{3}+\cdots=-(1-\frac{1}{2}+\frac{1}{3}-\cdots)=-\log 2$$2012-05-17
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    @Argon: You seem to miss the point I was trying to make. It is about justifying $\sum_{n=1}^{\infty} (a_n - b_n) = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{\infty} b_n$. For instance $0 = \sum_{n=1}^{\infty} (1/n - 1/n) = \sum_{n=1}^{\infty} 1/n - \sum_{n=1}^{\infty} 1/n$ might be meaningless. Of course, that is just my guess as to why you got a downvote. In your case, both $\sum a_n$ and $\sum b_n$ converge so it is not a problem.2012-05-17
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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.