Convergence of power series $\sum_{n=2}^{\infty} a^{\log_en}$

Question

Let $a>0$, the series

$$ \sum_{n=2}^{\infty} a^{\log_en} $$

is convergent for what values of $a$ ?

My attempt:

The usual way I think is to check $\lim_{n\rightarrow \infty} \frac{u_{n+1}}{u_n} < 1$ where, $u_n = a^{\log_en}$ and get condition on $a$ This leads to evaluating $\lim_{n \rightarrow \infty } a^{\log_e \frac{n+1}{n}}$ which is $1$. Right?

But, As we can see for $a=\frac{1}{e} , u_n = (\frac{1}{e})^{\log_en} = -n$ hence, the series diverges.

So, convergence for series happens for $0<a<\frac{1}{e}$, can anyone help me how to arrive at this conclusion using some theorem/result. Any hint will be appreciated ...

Answer 1

HINT:

Note that we have

$$a^{\log(n)}=n^{\log(a)}$$

Now, use the $p$ test.