Subsequence - absolute convergence

Alexmahone

Active member
Let $\{a_n\}$ be a sequence, and $\{a_{n_i}\}$ be any subsequence. Prove that if $\sum_{n=0}^\infty a_n$ is absolutely convergent, then $\sum_{i=0}^\infty a_{n_i}$ is absolutely convergent.

My attempt:

$\sum |\ a_n|$ is convergent.

$b_n=\left\{ \begin{array}{rcl}|a_{n_i}|\ &\text{for}& \ n=n_i \\ 0\ &\text{for}& \ n\neq n_i\end{array} \right.$

$0\le b_n\le\ |a_n|$ for all $n$.

Since $\sum |\ a_n|$ converges, $\sum b_n$ converges.

So, $\sum|\ a_{n_i}|$ converges. ($\because\sum_{n=0}^{n_i}b_n=\sum_{i=0}^i|\ a_{n_i}|$)

Is that okay?

Plato

Well-known member
MHB Math Helper
Let $\{a_n\}$ be a sequence, and $\{a_{n_i}\}$ be any subsequence. Prove that if $\sum_{n=0}^\infty a_n$ is absolutely convergent, then $\sum_{i=0}^\infty a_{n_i}$ is absolutely convergent.

My attempt:

$\sum |\ a_n|$ is convergent.

$b_n=\left\{ \begin{array}{rcl}|a_{n_i}|\ &\text{for}& \ n=n_i \\ 0\ &\text{for}& \ n\neq n_i\end{array} \right.$

$0\le b_n\le\ |a_n|$ for all $n$.

Since $\sum |\ a_n|$ converges, $\sum b_n$ converges.

So, $\sum|\ a_{n_i}|$ converges. ($\because\sum_{n=0}^{n_i}b_n=\sum_{i=0}^i|\ a_{n_i}|$)

Is that okay?
Yes, that works.