Proving Convergence of $\sum_{k=1}^{\infty} \frac{1}{k \, (\log (k+1))^p}$

[devious_]

Find all p \geq 0 such that

\sum_{k=1}^{\infty} \frac{1}{k \, (\log (k+1))^p}

converges.

It looks like the integral test is the most likely candidate, but I haven't been able to make any progress using it. I'd appreciate a push in the right direction.

Edit:
I've managed to prove that it converges for p > 1. Since it obviously diverges for p=0, I'm trying to see what happens when 0 < p \leq 1.

Edit2:
And now I just proved that it diverges for such p. Problem solved.

 

[quasar987]

out of curiosity, how did u proceed? I used a criterion that is very useful when dealing with log series. it says that \sum a_n converges \Leftrightarrow \sum 2^n a_{2^n} converges.

So compare the 2^n serie with the riemann p-serie and you get that the original series behaves just like the riemann p-serie, i.e. diverges for p \leq 1 and converges for p>1.

 

[matt grime]

How on Earth does that test work?

take the sum of (-1)^n/log(n), that converges by the alternating series test, yet the 2^n subseries

2^n(-1)^(2^n)/log(2^n) = 2^n/nlog(2)

does not converge.

 

[shmoe]

That's sometimes called the Cauchy-Condensation test:

http://mathworld.wolfram.com/CauchyCondensationTest.html

the terms must be nonnegative.

 

[quasar987]

Oh yeah, an must be decreasing non-negative.

 

[shmoe]

I was just about to add "non-increasing".

For the problem here, you can also use the integral test.

 

[devious_]

I used the fact that log is increasing so that \log (k) < \log (k+1), and I used the integral test.