Alternating series (Leibniz criterion)

[Damidami]

I read that an alternating series (http://en.wikipedia.org/wiki/Alternating_series_test) \Sigma (-1)^n a_n converges if "and only if" the sequence a_n is both monotonous and converges to zero.

I tried with this series:

\Sigma_{n=1}^{\infty} (-1)^n | \frac{1}{n^2} \sin(n)|

in the wolfram alpha (http://www.wolframalpha.com/input/?i=sum+%28-1%29^n+*+abs%28+1%2Fn^2+*+sin%28n%29+%29+from+n%3D 1+to+n%3D100) and seems to converge to -0.61..., even if a_n = |\frac{1}{n^2} \sin(n)| is not monotonous decreasing.

What am I doing wrong? Is the monotone condition necesary for this test, but the fact that a_n is not monotonous does not guarantee if the series converges or not?

Thanks.

[Office_Shredder]

The conditions given are sufficient, not necessary. It is possible for the terms to not be decreasing in magnitude, and for the whole series to still converge.

For your particular series, you can prove it converges absolutely by comparison with 1/n2

[Damidami]

The conditions given are sufficient, not necessary. It is possible for the terms to not be decreasing in magnitude, and for the whole series to still converge.

For your particular series, you can prove it converges absolutely by comparison with 1/n2

You are right, thanks! I confused because in class we saw this statement:

Let a_k > 0 be a sequence monotonous decreasing. Then \Sigma_{n=1}^{\infty} (-1)^k a_k converges \Leftrightarrow a_k \to 0

The \Rightarrow part of the proof is simply the necesary condition for any series to converge that a_k \to 0 .

But I thoght the \Leftarrow part would imply everything besides it: both that the series converges and that a_k is monotonous decreasing.

So it's a logical question now: when we have a double implication like above, the "leftarrow" implication only implies what the thesis says, but the hipothesis is the same. Am I right now?

Thanks!