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I want to show that $\sum_{n=1}^{\infty}\frac{(-1)^n}{n+1}$ converges by using that theorem.

Theorem:
suppose
(a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence
(b) $\dots \le b_2 \le b_1 \le b_0$
(c) $\lim_{n\to\infty}b_n=0$
then $\sum a_n b_n$ converges.

Put $a_n=(-1)^n$ , $b_n=\frac{1}{n+1}$
I think other things are okay but not sure about $A_n$ is a bounded sequence.
Or I'm wrong from the beginning? This theorem is not suitable for this proof?

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    $\begingroup$ yes you are right if $n = even$ $A_n$ is $0$ when $n=odd$ $A_n=1$ as result $A_n$ is bounded $\endgroup$ – jim Nov 26 '12 at 14:25
  • $\begingroup$ It's bounded because it can only receive two values: 0 and -1. $\endgroup$ – josh Nov 26 '12 at 14:30
  • $\begingroup$ The sequence {$\sum a_n$}={-1,1,-1,1,...} then $A_n$=0,-1,0,-1,... so I think it could be a divergence form when n is infinity and not bounded. $\endgroup$ – niagara Nov 26 '12 at 14:30
  • $\begingroup$ Oh, I just confused the definition of $A_n$. Now I understand. Thanks! $\endgroup$ – niagara Nov 26 '12 at 14:32
  • $\begingroup$ Funny: I think $\lim_{n\to \infty}=\sum_{n=0}^{\infty} (-1)^n $ is in fact divergent, it lacks a sum in the usual sense.... $\endgroup$ – draks ... Nov 26 '12 at 15:19

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