I'm practicing for my final exams this week but the past year papers have no answers so I'm not sure if my answers are acceptable, was hoping someone would look at my proof and let me know if it is a sufficient one, thank you in advance!

Problem : Suppose $\sum_{n=1}^{\infty} a_n$ converges absolutely. Let {$b_n$} be a subsequence of {$a_n$}. Show that $\sum_{n=1}^{\infty} b_n$ converges.

My Proof (1st attempt) : Since $\sum_{n=1}^{\infty} a_n$ converges absolutely, the sequence {$a_n$} is absolutely convergent as well. Any subsequence of an absolutely convergent sequence is also absolutely convergent, this implies that {$b_n$} is also an absolutely convergent sequence. Let {$b_n$} be a sequence of partial sum of $\sum_{n=1}^{\infty} b_n$. By definition of a Series, since the sequence of partial sum is convergent, the series $\sum_{n=1}^{\infty} b_n$ is convergent as well.

The issues I have with my 1st attempt is that 'absolutely convergent sequence' is quite a controversial phrase to me since I've seen many discussions on other posts stating that this does not exist or that it only exists on certain conditions. So, ultimately, I do not know if this is a valid condition to use in this problem. In addition, I do not know if I am able to let {$b_n$} be the sequence of partial sum of the series.

Thus, this brings me to my 2nd attempt at proving this :

2nd attempt : Since $\sum_{n=1}^{\infty} a_n$ is absolutely convergent, this means that the sequence of partial sums {$S_k$} is also convergent. Thus, {$S_k$} is a subsequence of {$a_n$}. Since {$b_n$} is also a subsequence of {$a_n$}, by the Bolzano-Weierstrass Theorem, {$b_n$} is also convergent. Let {$S_l$} be a subsequence of {$b_n$}, and consider $\sum_{n=1}^{\infty} b_n$. Let {$S_l$} be the sequence of partial sums of $\sum_{n=1}^{\infty} b_n$. Thus, by the definition of series, since the sequence of partial sums is convergent, $\sum_{n=1}^{\infty} b_n$ is convergent as well.

The problems I currently have with this 2nd proof is that I'm not sure if I'm applying the Bolzano-Weierstrass theorem properly. In addition, stating that {$b_n$} is also convergent seems a little forced to me but I have no idea how to put it in a more convincing statement.

Somehow I understand all the theorems but I have no idea how to put them all together to make a solid proof for this question.

Hope someone can help me out with this, thank you.

  • 1
    $\begingroup$ $\sum |b_n| \leq \sum |a_n| < \infty$ might be useful to observe. $\endgroup$
    – Atbey
    Commented Jun 20, 2018 at 14:34
  • $\begingroup$ Atbey.Want to write a short answer? $\endgroup$ Commented Jun 20, 2018 at 14:48
  • $\begingroup$ @Atbey does that mean the partial sums of {$b_n$} is less than partial sums of {$a_n$} because it is a subsequence? But how would this help prove the Series is convergent though? $\endgroup$ Commented Jun 20, 2018 at 14:53
  • $\begingroup$ I do not have access to computer @PeterSzilas. Absolutely convergent sequence is convergent is equivalent to space being complete. $\endgroup$
    – Atbey
    Commented Jun 20, 2018 at 15:15

3 Answers 3


I assume that you mean a series of real or complex numbers. The Cauchy convergence criterion says that $\Sigma_{n=1}^\infty c_n$ converges if and only if for each $\epsilon > 0$ there exists $m \in \mathbb{N}$ such that for all $m' \ge m$

$$\lvert \Sigma_{n=m}^{m'} c_n \rvert < \epsilon .$$

Now your series $\Sigma_{n=1}^\infty a_n$ converges absolutely, i.e. $\Sigma_{n=1}^\infty \lvert a_n \rvert$ converges. Let $(b_n)$ be a subsequence of $(a_n)$. This means you have $b_n = a_{\varphi(n)}$ with a strictly increasing function $\varphi : \mathbb{N} \to \mathbb{N}$. Let $\epsilon > 0$. You find $m \in \mathbb{N}$ such that for all $m' \ge m$

$$\Sigma_{n=m}^{m'} \lvert a_n \rvert = \lvert \Sigma_{n=m}^{m'} \lvert a_n \rvert \rvert < \epsilon .$$


$$ \lvert \Sigma_{n=m}^{m'} \lvert b_n \rvert \rvert = \Sigma_{n=m}^{m'} \lvert b_n \rvert = \Sigma_{n=m}^{m'} \lvert a_{\varphi(n)} \rvert \le \Sigma_{n=\varphi(m)}^{\varphi(m')} \lvert a_n \rvert \le \Sigma_{n=m}^{\varphi(m')} \lvert a_n \rvert < \epsilon .$$

The first $\le$ comes from the fact all the summands $\lvert a_{\varphi(n)} \rvert$ with $n = \varphi(m), ..., \varphi(m')$ occur as suitable summands of the form $\lvert a_n \rvert$ with $n = \varphi(m), ...., \varphi(m')$. The second $\le$ is due to the fact that $m \le \varphi(m)$.


Correct me if wrong:

$S_n =\sum_{k=1}^{n}|b_k| \le \sum_{k=1}^{\infty}|a_k| =: M$, real, positive.

1) $S_n$ is bounded above by $M.$

2) $S_n$ is monotonically increasing.

It follows that $S_n$ is convergent.

3)Convergence of $ \sum_{k=1}^{\infty}|b_k|$ implies the

convergence of $\sum_{k=1}^{\infty}b_k.$

Note: To prove 3) use Cauchy criterion, and triangle inequality.


If $\sum a_n$ converges then the sequence $(a_n)_n$ converges to $0.$ The rest of both of your attempts is "hand-waving". A marker might give you a $0.$

Let $\sum a_n$ be absolutely convergent. Let $f:\Bbb N\to \Bbb N$ be strictly increasing and let $b_n=a_{f(n)}.$

Given $\epsilon >0,$ take $m_1\in \Bbb N$ such that $m_1<n_1\leq n_2\implies \sum_{n=n_1}^{n_2}|a_n|<\epsilon.$

Whenever $n_3\leq n_4$ we have $$\{f(n):n_3\leq n\leq n_4\}\subset \{n\in \Bbb N: f(n_3)\leq n\leq f(n_4)\}.$$ Take $m_2$ such that $f(m_2)>m_1.$ Then $$m_2<n_3\leq n_4\implies \epsilon >\sum_{n=f(n_3)}^{f(n_4)}|a_n|\geq \sum_{n=n_3}^{n_4}|a_{f(n)}|=\sum_{n=n_3}^{n_4}|b_n|.$$ So the Cauchy Criterion is met for $\sum_n|b_n|.$


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