
Series convergence
I have two series:
$\displaystyle \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.
$\displaystyle \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? :(

Re: Series convergence
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 $\displaystyle \lim_{n \to \infty} {2 \ln n \over (n+1)^{0.5}}$?

Re: Series convergence
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.

Re: Series convergence
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?

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

Re: Series convergence
I'm guessing that's a no?