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?
 A: 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
$$a_n = \begin{cases} k^{-2} &, n = n_k \\ 2^{-n} &, n \notin \{ n_k : k \in \mathbb{N}\}.\end{cases}$$
Then it's easily seen that $\sum a_n$ converges, yet we have
$$(a_{n_k})^{3/2}\cdot \frac{\exp (a_{n_k}\ln n_k) - 1}{a_{n_k}\ln n_k} > k^{-3}\cdot \frac{a_{n_k} \ln n_k}{2} = \frac{1}{2} k^{-5} \ln n_k > \frac{4^k}{2k^5}$$
since $\frac{e^x-1}{x} > \frac{x}{2}$ for $x > 0$, and we see that the terms in $(1)$ are unbounded, whence $(1)$ diverges for this choice of $(a_n)$.
A: Taking the term in $x^2$ in the power series for $e^x,$ we have $e^x-1>x^2/2$ when $x>0.$
It can be true that for any $k>0,$ the set $S(k)=\{ n: a_n>(2k/\log n)^{2/5}\}$ is infinite. For  $n\in S(k)$ we have $n^{a_n}-1=e^{a_n \log n}-1>a_n^2(\log n)^2/2$ and $$\frac {\sqrt a_n}{\log n} (n^{a_n}-1)> \frac {\sqrt a_n}{\log n}a_n^2(\log n)^2/2=\frac {1}{2}a_n^{5/2}\log n>k .$$ 
Q: What if $(a_n)_n$ is a monotonic sequence?
A: 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
$$\lim\limits_{n\to\infty} a_{n}{\ln\, n}=0
$$
