TITLE: Prove that $\sum a_n$ converges $\implies \sum \frac {a_n} {2^n}$ converges.
QUESTION [5 upvotes]: Prove that $\sum a_n$ converges $\implies \sum a_n/2^n$ converges.
My proof is below.  Please verify or improve it.
$\lim a_n = 0$, as $\sum a_n$ converges.  Therefore, there exists $N$ such that for all $n > N$, $|a_n| < 1$.  For $n > N, 0 \leq |a_n/2^n| < 1/2^n$.  For any series $p_n, q_n$,  if $\sum q_n$ converges, and for all $n > N$, $0 \leq p_n \leq q_n$, then $p_n$ converges; consequently, $\sum |a_n/2^n|$ converges.  And since for any series $p_n$, if $\sum |p_n|$ converges, then $\sum p_n$ converges, therefore $\sum a_n/2^n$ converges.  QED.

Update: Weaker Conditions
The condition $\sum a_n$ converges is unnecessarily strong. The proof works as long as $a_n$ eventually becomes bounded.

REPLY [3 votes]: Another proof: use the root test. We have
$\limsup\limits_{n \to \infty} \sqrt[n]{|a_n|} \leq 1$
Since $\sum\limits_{n = 0}^\infty a_n$ converges. Thus, we have
$\limsup\limits_{n \to \infty} \sqrt[n]{|\frac{a_n}{2^n}|} = \limsup\limits_{n \to \infty} \sqrt[n]{\frac{|a_n|}{2^n}} = \limsup\limits_{n \to \infty} \frac{1}{2} \sqrt[n]{|a_n|} = \frac{1}{2} \limsup\limits_{n \to \infty} \sqrt[n]{|a_n|} \leq \frac{1}{2} \cdot 1 = \frac{1}{2} < 1$
So $\sum\limits_{n = 0}^\infty \frac{a_n}{2^n}$ converges.
Note: you cannot use the ratio test here because it may be that $\lim\limits_{n \to \infty} \frac{a_{n + 1}}{a_n}$ is undefined. But you can use the root test because $\limsup\limits_{n \to \infty} \sqrt[n]{|a_n|}$ always exists or is infinity.