real analysis - Convergence of the series $sum limits_{n=2}^{infty} frac{1}{nlog^s n}$

We all know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ converges for $s>1$ and diverges for $s \leq 1$ (Assume $s \in \mathbb{R}$).

I was curious to see till what extent I can push the denominator so that it will still diverges.

So I took $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges).

No surprises here. I expected it to diverge since $\log n$ grows slowly than any power of $n$.

However, when I take $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}$, I find that it converges $\forall s>1$.

(By the same argument as previous).

This doesn't make sense to me though.

If this were to converge, then I should be able to find a $s_1 > 1$ such that

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$

Doesn't this mean that in some sense $\log n$ grows faster than a power of $n$?

(or)

How should I make sense of (or) interpret this result?

(I am assuming that my convergence and divergence conclusions are right).

Answer

Yes $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^s}$ is convergent if $\displaystyle s > 1$, we can see that by comparing with the corresponding integral.

As to your other question, if

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$

does not imply $\log n$ grows faster than a power of $\displaystyle n$. You cannot compare them term by term.

What happens is that the first "few" terms of the series dominate (the remainder goes to 0). For a small enough $\displaystyle \epsilon$, we have that $\log n > n^{\epsilon}$ for a sufficient number of (initial) terms, enough for the series without $\log n$ to dominate the other.