Series convergence

**Lisa91** · January 15th 2013, 12:32 PM

I have two series:
$\sum_{n=13}^{\infty} (-1)^{\left\lfloor \frac{n}{13} \right\rfloor} \frac{lnn}{nln(lnn)}$ . I wanted to 'group' terms but I had a problem with idices.

$\sum_{n=1}^{\infty} (-1)^{n} \frac{2lnn}{(n+1)^{0,5}}$ I tried the Leibniz's, Dirichlet's, Abel's tests but they gave me nothing. Any hints how to cope with them?

**ILikeSerena** · January 15th 2013, 01:16 PM

Hi Lisa91!

Let's start with the second.

You said you tried Leibniz and that it gave you nothing.
From the formula there's a pretty good match with Leibniz's test, so let's investigate.

Can you find $\lim_{n \to \infty} {2 \ln n \over (n+1)^{0.5}}$ ?

**Lisa91** · January 15th 2013, 01:24 PM

Well, I suppose it's 0 but I'm not so sure I know how to prove it. I wanted to show that the sequence declines but I couldn't finish it.

**ILikeSerena** · January 15th 2013, 01:30 PM

Yes, it is zero.
So if it is also decreasing, we can apply Leibniz.

Let's start with the first criterion.
Are you aware of l'Hôpital's rule?

**SworD** · January 15th 2013, 01:30 PM

The "power" of any log is 0, and the power of the denominator is 0.5, so the ratio will go to 0.

**ILikeSerena** · January 16th 2013, 02:05 PM

I'm guessing that's a no?

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