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 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.

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.