Inequality of $\limsup$ from convergence of a sum Let $(a_n)$ be a strictly increasing positive sequence with limit infinity. Suppose that $x > 0$ is given such that $\sum e^{-xa_n}$ converges. Is it true that $\limsup\limits_{n\to\infty} \frac{\log n}{a_n} \leq x$?
My attempt:
I think it's true, but can't prove it.
I can expand the definition of the limit superior, of course. However trying to prove by contradiction, I ran into a problem with comparison test, namely I have a subsequence $(n_p)$ for which $\frac{1}{n_p} < e^{-xa_{n_p}}$, with no way to control the sequence $(n_p)$ (and thus no way to conclude divergence of the sum).
I also tried using the fact that $e^{-xa_n} \to 0$, but got nowhere.
Help, please!
 A: Set $b_n := e^{-xa_n}$. Then $(b_n)$ decreases and tends to zero. We have $a_n = -x^{-1}\log(b_n)$ and thus
$$
\frac{\log(n)}{xa_n} = \frac{\log(n)}{\log(1/b_n)}.
$$
Now, $\limsup\frac{\log(n)}{\log(1/b_n)}\le 1$ means that for all $\delta > 0$ we have $\frac{\log(n)}{\log(1/b_n)}>1+\delta$ only for finitely many $n$. The latter inequality is equivalent to $b_n > n^{-1/(1+\delta)}$. Hence, we have to show that for all $p\in (0,1)$ there are at most finitely many $n$ such that $b_n > \frac 1{n^p}$.
Assume the contrary. Then there is a subsequence $(b_{n_k})$ with $b_{n_k} > \frac 1{n_k^p}$ for all $k$. From
\begin{align}
\sum_nb_n
&\ge \sum_{k=2}^\infty\sum_{n=n_{k-1}+1}^{n_k}b_n\,\ge\,\sum_{k=2}^\infty\sum_{n=n_{k-1}+1}^{n_k}b_{n_k} = \sum_{k=2}^\infty(n_k-n_{k-1})b_{n_k}\ge \sum_{k=2}^\infty\frac{n_k-n_{k-1}}{n_k^p}.
\end{align}
Now, for $K$ large we have
\begin{align}
\sum_{k=2}^K\frac{n_k-n_{k-1}}{n_k^p}
&= \sum_{k=2}^Kn_k^{1-p} - \sum_{k=1}^{K-1}\frac{n_{k}}{n_{k+1}^p}\\
&= n_K^{1-p} + \sum_{k=2}^{K-1}n_k^{1-p}\frac{n_{k+1}^p-n_k^p}{n_{k+1}^p} - \frac{n_1}{n_2^p}\\
&\ge\,n_K^{1-p} - \frac{n_1}{n_2^p}
\end{align}
Hence,
$$
\sum_nb_n\ge\lim_{K\to\infty}n_K^{1-p} = \infty.
$$
A contradiction!
