Does $\sum\frac{1}{nf(n) \log(n)}$ converge?

Question

Let $f$ be an unbounded non-decreasing function s.t. $\sum\frac{1}{2^{f(n)}}$ converges.

Does $\sum\frac{1}{nf(n) \log(n)}$ converge?

My thoughts:

Since $\sum\frac{1}{2^{f(n)}}$ converges, then $\ logn = O(f(n))$, otherwise it will diverge.

And now we can use comparison test to find an upper bound using $a_n = 1/n \log^2n$.

Am I correct?

thanks.

Presumably the sums start at $n=2$. – J.G. Jun 23 '22 at 15:02 — J.G., Jun 23 '22 at 15:02

score 3 · Accepted Answer · answered Jun 23 '22 at 14:01

The assumption that $f(n)$ is non decreasing is essential in order to show that $\log n=O(f(n)).$

The condition $\sum 2^{-f(n)}<\infty$ combined with the monotonicity assumption imply (see) $${n\over 2^{f(n)}}\to 0$$ Hence $$\log n =o(f(n))$$ and the conclusion follows by the comparison test, as observed by OP.

Does $\sum\frac{1}{nf(n) \log(n)}$ converge?

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