If sequence $a_n>0$ is decreasing to $0$ and $\sum_{k=1}^\infty a_n=\infty$, prove $\sum_{n=1}^\infty\frac{a_n}{e^{S_n}}<\infty$ If sequence $a_n>0$ is decreasing to $0$ for all $n\geq 1$ and $\sum_{n=1}^\infty a_n=\infty$, is it always true that $\sum_{n=1}^\infty\frac{a_n}{e^{S_n}}<\infty$, where $S_n=\sum_{k=1}^na_k$ is the partial sum? If yes, how to prove; if no, is there any counterexample?

I have tried $a_n=\frac{1}{n}$, in that case $e^{S_n}\sim n$ when $n$ is large, so $\sum_{n=1}^\infty \frac{a_n}{e^{S_n}} \sim \sum_{n=1}^\infty \frac{1}{n^2}<\infty$. I also tried $a_n=\frac{1}{n\ln n}$, in this case $e^{S_n}\sim \ln n$ when $n$ is large, so $\sum_{n=1}^\infty \frac{a_n}{e^{S_n}} \sim \sum_{n=1}^\infty \frac{1}{n(\ln n)^2}<\infty$ by Cauchy condensation test.

The only condition needed is that $a_n>0$ and then $\sum_{n \ge 1} a_ne^{-S_n} \le 1$ since with $S_0=0<S_1<S_2<...$ one has $$e^{-S_n} \le \frac{\int_{S_{n-1}}^{S_n}e^{-x}dx}{a_n}, n \ge 1$$
Then $\sum_{n=1}^N a_ne^{-S_n} \le \int_0^{S_N}e^{-x}dx \le 1$
