In posting this question, I noticed a lot of 'similar' threads pop up, but felt that they required a fundamentally different approach.

If any of you feel differently, please feel free to vote this thread as a duplicate and I will delete it.

Here is my approach, I would appreciate help/correction where relevant, as I am unsure of how robust my answer is:

If $\sum a_n$ converges absolutely, then we have that $\sum |a_n|$ converges. This implies that $\sum a_n$ also converges, and that the sequence $(a_n)_{n \in \mathbb{N}}\to 0$.$^{(1)}$

This means that the sequence $(a_n)$ is bounded and monotone (decreasing), and as such convergent.

Looking at $(a_{2n})$ we notice it is a sub-sequence of $(a_n)$ and converges to the same value (Bolzano-Weiestrass).$^{(2)}$

Then the partial sums $(s_{2n}) \to A$ and thus $\sum a_{2n} = A$

Is this sufficient? I feel it is a little wishy/washy in using bolzano-weistrass.

Do I need to formally show $(a_{2n})$ is a sub-sequence of $(a_n)$

Any tips/hints/corrections are greatly appreciated.

$^{(1),}$$^{(2)}$ - I am not required to show these results, as quoting the theorem from my notes is considered sufficient by my professor.

  • 1
    $\begingroup$ Your use of Bolzano Weierstrass is useless in this context. Moreover, you are confusing $a_1 + a_2 + \dots + a_{2n} = \sum_{k=1}^{2n} a_k$ with $a_2 + a_4 + \dots + a_{2n} = \sum_{k=1}^n a_{2k}$: these two sequences (you are denoting both of them with the same symbol $s_{2n}$) are clearly different, so your argument makes no sense. $\endgroup$
    – Crostul
    Commented Apr 9, 2015 at 18:15
  • $\begingroup$ Thanks @Crostul I had a feeling this was the case (re: denoting the sequences incorrectly). I just want to clarify, instead of using $(s_{2n}) \to A$ to denote the partial sums of $\sum a_{2n}$, what should I have done? Let $b_n = a_{2n}$ and $(t_k)$ be the partial sums of $b_n$ ? $\endgroup$
    – elbarto
    Commented Apr 9, 2015 at 19:35

1 Answer 1


$$ \sum_{n\in \mathbb{N}} |a_{2n}| \leq \sum_{n\in \mathbb{N}}|a_n| < \infty$$

  • $\begingroup$ I like this idea, it is much more elegant. Would you say there were any errors in my reasoning, however? $\endgroup$
    – elbarto
    Commented Apr 9, 2015 at 17:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .