Suppose $\sum a_n $ converges. Does $\sum 2^{-n} a_n $ always converge? Suppose $\sum a_n $ converges. Does $\sum 2^{-n} a_n $ always converge ? 
If $a_n$ is non-negative series then the answer is yes. 
What about the general series? 
 A: Indeed, the assumption that $\sum_n a_n$ converges is useless. The conclusion remains true for any bounded sequence $\{a_n\}_n$. For if $\sup_n |a_n| = M$, then
$$
\sum_n \left| \frac{a_n}{2^n} \right| \leq M \sum_n \frac{1}{2^n} < +\infty.
$$
More generally, the abolute convergence of $\sum_n a_n b_n$ is guaranteed as soon as $\{a_n\}_n$ is bounded and $\sum_n |b_n| < +\infty$.
A: I think the answer is yes in the genral case because : 
if $\sum a_n$ converge then $a_n \to  0$ then there is $n_0$ such that for every $n\ge n_0$ $|a_n|\le1$ and than we can use the comparastion test , and show that the new sequence is converge. am i right ?
A: We know from assumption of convergence that for every $\epsilon>0$ there exists an $N^*\in \mathbb N$ such that $|a_n|<\epsilon$ for all $n \geq N^*$.
Since $|2^{-n}a_n| \leq |a_n|$ for all $n \in \mathbb N$, we can select the same $N^* \in \mathbb N$ we used in the previous paragraph to ensure that $|2^{-n}a_n|<\epsilon$ for all $n\geq N^*$ and conclude that $\sum_n 2^{-n}a_n$ is convergent.
