# If $\sum_{n}A_n$ is convergent, then $\sum_{n} (-1)^{n} A_n$ is also convergent?

If you know that the series $\sum_{n}A_n$ is convergent, i have to prove or give a counterexample of the following statement:

The series $\sum_{n} (-1)^{n} A_n$ is also convergent.

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The answers below are good. An additional thing to think about: If $A_n$ is monotone and converges to 0, then is $\sum (-1)^n A_n$ convergent? –  mez Feb 16 '13 at 12:37
does my answer help you? if so you can accept it –  Dominic Michaelis Mar 21 '13 at 20:38

Take $$a_n=(-1)^n \frac{1}{n}$$ It is known that $$\sum_{n=1}^\infty (-1)^n \frac{1}{n}=-\log(2),$$ but $$\sum_{n=1}^\infty (-1)^n (-1)^n \frac{1}{n}=\sum_{n=1}^\infty \frac{1}{n}$$ diverges. If you have $A_n\geq 0$ for nearly all $n\in \mathbb{N}$ you have the absolute konvergence and then the statement is true

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Thank you, but how do you know that the second sum is - log 2? –  Sjoerd Smaal Feb 16 '13 at 10:25
it's from the taylor series of the logarithm, if you just need the convergence take leibniz test –  Dominic Michaelis Feb 16 '13 at 10:26

It's false, take $$A_n = (-1)^n\frac{1}{n}$$. But if $A_n \geq 0$ for all $n$ then it's true by the Leibniz's condition for alternating series.

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If $A_n \ge 0$, it is true, but not by Leibniz's test. (The sequence $A_n$ doesn't have to be decreasing.) It is true by absolute convergence, though. –  mrf Feb 16 '13 at 11:20
Yes you are right. –  Diego Silvera Feb 16 '13 at 11:26