If $\sum_{n=2}^{\infty} a_n$ is convergent, is $\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)$ convergent as well? Suppose $\sum_{n=2}^{\infty} a_n$ is a convergent series. Is the following series convergent as well?
$$\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)$$
The part with $n^{a_n}-1 $ looks quite similar to $\sqrt[n]{n}-1$ which behaves like $\ln{n}$, however, I'm not able to tie this fact to the behavior of the whole expression. Any suggestions?
We can rewrite $n^{a_n}$ as $\exp (a_n \ln n)$, and then

$$\sum_{n = 2}^{\infty} \frac{\sqrt{a_n}}{\ln n}\bigl(n^{a_n} - 1\bigr) = \sum_{n = 2}^{\infty} a_n^{3/2}\cdot \frac{\exp (a_n\ln n) - 1}{a_n\ln n}.\tag{1}$$

Now, assuming $a_n \geqslant 0$, everything is fine and dandy if $a_n\ln n$ remains bounded. Of course, if a series of nonnegative terms converges, the terms must on average decay much faster than $\frac{1}{\ln n}$, so it may look as though we can deduce convergence. But that's only on average, and for sparse enough indices, we can have arbitrarily large $a_n\ln n$. So large that the series in $(1)$ diverges. For $k \in \mathbb{N}$, let $n_k = \lceil \exp (4^k)\rceil$, and

$$a_n = \begin{cases} k^{-2} &amp;, n = n_k \\ 2^{-n} &amp;, n \notin \{ n_k : k \in \mathbb{N}\}.\end{cases}$$

Then it's easily seen that $\sum a_n$ converges, yet we have

$$(a_{n_k})^{3/2}\cdot \frac{\exp (a_{n_k}\ln n_k) - 1}{a_{n_k}\ln n_k} &gt; k^{-3}\cdot \frac{a_{n_k} \ln n_k}{2} = \frac{1}{2} k^{-5} \ln n_k &gt; \frac{4^k}{2k^5}$$

since $\frac{e^x-1}{x} &gt; \frac{x}{2}$ for $x &gt; 0$, and we see that the terms in $(1)$ are unbounded, whence $(1)$ diverges for this choice of $(a_n)$.
For large values of $n$,

$$n^{a_n}-1={1+a_n{\ln\, n}}-1 =a_n{\ln\, n},$$

so the given series becomes

$$\sum_{n=2}^{\infty}\frac{\sqrt{a_{n}}}{\ln\, n}(n^{a_{n}}-1)=\sum_{n=2}^{\infty}\frac{\sqrt{a_{n}}}{\ln\, n}(a_n){\ln\, n} = \sum_{n=2}^{\infty}{a_n}^\frac{3}{2},$$

which converges , if
$$\lim\limits_{n\to\infty} a_{n}{\ln\, n}=0
$$