I am trying to figure out if the following series converges or diverges:

Let $a_n$ be an unbounded non decreasing sequence s.t. $\sum \frac{a_n}{n^a}$ diverges, for $a > 2$.

Does $\sum 1/a_n$ converge?

I wanted to prove the statement using the comparison test with $b_n = \frac{1}{n^{1 + \epsilon}}$ for some positive $\epsilon$ and show $\frac{a_n}{b_n}$ must converge but could not go any further since i was not able to prove such $b_n$ exists.

Any hints will be appreciated.


1 Answer 1


Its false, here is a counterexample. We define $a_n$ inductively. Start with $a_0=1$ and start with auxiliary sequence $k(n)$ with $k(0)=0$. Now:

If $k(n-1)=0$ set $a_n=2^n$ and $k(n)=2^n$. Else set $a_n=a_{n-1}$ and $k(n)=k(n-1)-1$.

Note that $k(n)=0$ will happen for infinitely many $n$, and for any $n$ where this happens you have $a_n=2^n$, so $\frac{a_n}{n^a}$ will be unboundedly large for such $n$, in particular the series over this has no hope of converging.

On the other hand if $n$ is so that $k(n)=0$ you have for all $m\in[n, n+2^n]$ that $a_m=2^n$, so: $$\sum_{m=n}^{n+2^n}\frac1{a_m}=2^n\cdot\frac1{2^n}=1$$ and you see that also the series $\sum \frac{1}{a_n}$ cannot converge.


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