3
$\begingroup$

Given $$(a_1+a_2)+(a_3+a_4)+(a_5+a_6)+(a_7+a_8)+\cdots= S$$ if $a_n\to 0$, then $$a_1+a_2+a_3+a_4+a_4+a_5+a_6+a_7+\cdots= S$$

My attempt: The sequence of partial sums of $\sum a_n$ is $$(s_n)=(a_1,\ a_1+a_2,\ a_1+a_2+a_3,\ a_1+a_2+a_3+a_4\ ...)$$ Witch has a subsequence $(s_{2n})=(a_1+a_2,\ a_1+a_2+a_3+a_3,\ ...)$ that is the sequence of partial sums to $\sum(a_n+a_{n+1})$. This implies that it converges to $S$.

Also, there is the subsequence $(s_{2n+1})=(a_1,\ a_1+a_2+a_3,\ a_1+a_2+a_3+a_4+a_5\ \cdots)$ that converges to $\lim s_{2n+1}=\lim [ \sum(a_n+a_{n+1})+a_{2n+1}]=\lim\sum(a_n+a_{n+1})+\lim a_{2n+1}=S+0$

Since $\lim s_{2n}=\lim s_{2n+1}=S\implies \lim s_n=S$


Is this correct? Any tips on the solution? Thanks!

$\endgroup$
  • 1
    $\begingroup$ It looks right, but I think the last line of the argument needs a little more justification. $\endgroup$ – spaceisdarkgreen Jul 28 '17 at 18:47
  • 1
    $\begingroup$ @Br.M: There's a relatively standard exercise that if real sequences $(a_{k})$ and $(b_{k})$ converge to the same limit $S$, then the "shuffled" sequence $a_{1}, b_{1}, a_{2}, b_{2}, \dots$ converges to $S$. If you have this result, you're pretty much done, depending how careful you need to be with $\varepsilon$s and $N$s. (Separately, the title question does not match the body question, because the "terms approach zero" hypothesis is omitted from the title.) $\endgroup$ – Andrew D. Hwang Jul 28 '17 at 18:59
  • $\begingroup$ Why is $a_4$ the only one repeated twice? $\endgroup$ – Aryabhata Jul 29 '17 at 0:44
  • $\begingroup$ Oh, it's a typo... sorry $\endgroup$ – B. de Morais Jul 29 '17 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.