Prove that if $\sum a_n$ converges absolutely, then $\sum a_{2n}$ converges.
$$ \sum_{n\in \mathbb{N}} |a_{2n}| \leq \sum_{n\in \mathbb{N}}|a_n| < \infty$$
Related videos on Youtube
elbarto
Updated on August 02, 2022Comments
-
elbarto over 1 year
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.
-
Crostul over 8 yearsYour 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.
-
elbarto over 8 yearsThanks @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 $b_n$ ?
-
-
elbarto over 8 yearsI like this idea, it is much more elegant. Would you say there were any errors in my reasoning, however?