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Suppose $\sum_{n = 1}^{\infty}a_{n}^{2}$ converges. Is it necessarily true that $\sum_{n = 1}^{\infty}a_{n}e^{-n}$ converges?

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Since $\sum a_n^2$ converges, there exists $N$ such that for all $n > N$, we have $\vert a_n \vert < \epsilon$. Hence, beyond this $N$, we have $\sum_{n >N} a_n e^{-n}$ bounded by a geometric series i.e. $\left \vert \sum_{n \geq N} a_n e^{-n} \right \vert \leq \epsilon \sum_{n \geq N} e^{-n} = \epsilon \dfrac{e^{-N}}{1-e^{-1}}$ Hence, $\displaystyle \sum_{n \geq 1} a_n e^{-n}$ converges.

All you need is $a_n \to 0$ as $n \to \infty$, you do not even need $\sum a_n^2$ to converge.

EDIT

As Joel Cohen points out, it is enough that $a_n$ is bounded.

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    Actually, even $(a_n)$ is bounded would suffice in your proof.2012-10-31