Let $p,q \in \mathbb{R}$
Consider :
$$\sum_{n=2}^\infty \frac{1}{n^p(\ln(n))^q}$$
For what $p,q$ does the series converge ?
Here's what I've got.
It suffices to check : $$\sum_{n=2}^\infty \frac{2^n}{(2^n)^p(\ln(2^n))^q} =\sum_{n=2}^\infty \frac{(2^{1-p})^n}{n^q} $$
So we require $ p>1$ or else the limit isnt 0 and it clearly diverges.
Now assuming $p>1 \implies |2^{1-p}| = z<1$
So by the Ratio Test $$\sum_{n=2}^\infty \frac{z^n}{n^q} $$
Converges $\forall q$
Hence the series converges if $p >1$
Is this right?
Best Answer
The series converges if $p > 1$, or if $p = 1$ and $q > 1$. It is possible to tackle both statements using condensation criteria, like you used (perhaps under a different name) in your proof.
It is also possible (perhaps even advisable) to use the integral test. So you are left trying to understand when $$ \int_2^\infty \frac{1}{x^p \ln^q x} dx $$ converges. The $p > 1$ case can be handled by direct comparison with $\dfrac{1}{x^p}$. The $p = 1$ case is where the integral test really shines.