MHB Subsequence - absolute convergence

[alexmahone]

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?

 

[Plato2]

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.

 

[Greg Bernhardt]

Yes, your proof is correct. You have correctly used the fact that the absolute value of a subsequence is always less than or equal to the absolute value of the original sequence, and that the convergence of the original series implies the convergence of the subsequence. Well done!