# 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.
The text passage contains a mathematical question and provides a detailed answer to it, including multiple approaches and justifications.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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.
user
The text passage contains a mathematical question and provides a detailed answer to it, including multiple approaches and justifications.

TYPE: SINGLE_QUESTION_WITH_ANSWER