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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.

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    yes, $e$ means euler constant2012-11-15

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Use Abel's identity to show that:

$\sum_{n=1}^{k}\frac{S_n}{e^n}=1/(e^{-1}-1)[e^{-(k+1)}S_{k+1}-e^{-1}S_1-\sum_{n=1}^{k}\frac{a_{n+1}}{e^{n+1}}]$

Now the answer to your question follows very easily because the limit of the right hand side exists.

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    +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.2012-11-14