If $\sum\limits_{k=1}^{\infty}a_k=S $, then $ a_4+a_3+a_2+a_1+a_8+a_7+a_6+a_5+\dots=?$

Question

if we know that $\sum\limits_{k=1}^{\infty}a_k=S$, what can we say about the convergence of $$a_4+a_3+a_2+a_1+a_8+a_7+a_6+a_5+a_{12}+a_{11}+a_{10}+a_{9}+\dots$$ ?

If it does converges, what is the sum (in terms of $S$)?

As per the first question - it clearly converges since the number of terms in each parentheses is bounded (by 4) and the $(a_n)_{n=1}^\infty$ tends to zero as $n\to\infty$.

Second question is where I'm struggling. We don't know that $\sum\limits_{k=1}^{\infty}a_k$ absolutely converges so I don't know what can we say about it's sum.

Thanks for your help.

Answer 1

I think Cameron is right. In particular, the difference $|S_n-T_n|$ is bounded by $|a_{n-1}+a_{n-2}+a_{n-3}|$ and therefore has to approach zero.

The problem of series that are not absolutely convergent is that you can't make arbitrary rearrangement of the terms. However, in general, the rearrangement that cause the sum of a convergent series to change cannot "bounded", meaning that there's no uniform upper bound on the number of shifts applied to each term of the original series (i.e. you can't say "every term is moved at most by $M$ terms").

Indeed, if every term of the sequence is shifted by at most $M$ terms, you can prove that the difference between the partial sums of the new and original series is bounded by

$$|T_n-S_n|\leq\sum_{n-M}^{n-1}|a_n|$$

which clearly converges to zero if the original series is convergent.

Answer 2

It should actually be $S$, too. Consider the respective sequences of partial sums, $S_n,T_n$, and show that $|S_n-T_n|$ converges to $0$.