# show that $\sum_{n=1}^{\infty}\frac{a_n}{e^n}$ implies $\sum_{n=1}^{\infty}\frac{S_n}{e^n}$ where $S_n=\sum_{k=1}^{n}a_n$

If $$\sum_{n=1}^{\infty}\frac{a_n}{e^n}$$ is convergent

denote$$S_n=\sum_{k=1}^{n}a_n$$

show that $$\sum_{n=1}^{\infty}\frac{S_n}{e^n}$$ is convergent.

-
Just to clarify, $e=\ln(1)$ here, right? Its a constant? – icurays1 Nov 14 '12 at 16:43
@icurays1 Rather $\ln e=1$. – Hagen von Eitzen Nov 14 '12 at 16:49
Ha, whoops. Coffee hasn't kicked in yet. Yes, that's what I meant. – icurays1 Nov 14 '12 at 16:51
yes, $e$ means euler constant – Laura Nov 15 '12 at 0:45

$$\sum_{n=1}^{k}\frac{S_n}{e^n}=1/(e^{-1}-1)[e^{-(k+1)}S_{k+1}-e^{-1}S_1-\sum_{n=1}^{k}\frac{a_{n+1}}{e^{n+1}}]$$
+1 for nice argument! I want to comment that though the convergence of $S_{k} e^{-k}$ follows from Cesaro-Stolz theorem together with the observation $a_n = o(e^n)$, it still seems to deserve a justification when it comes to elementary level analysis. – Sangchul Lee Nov 14 '12 at 16:53