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I have to use

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

to compute



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

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: $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$. – Did May 17 '12 at 20:46
up vote 3 down vote accepted

Noting that


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). – Aryabhata May 17 '12 at 21:04
@Aryabhata Which $-1$ are you referring to? – Argon May 17 '12 at 21:05
Someone downvoted this answer, and for a valid reason. – Aryabhata May 17 '12 at 21:13
@Argon: Thanks for your reply. Can you please verify the solution. – preeti May 17 '12 at 21:15
@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. – Aryabhata May 17 '12 at 21:41


Consider partial fraction decomposition.

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Thanks for the hint. – preeti May 17 '12 at 21:15

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