text: # 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.
  
  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 $\sum_n |b_n| < +\infty$.
  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 ?
  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.