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

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.

• Actually, the terms tending to $0$ isn't sufficient for series convergence on its own (consider the harmonic series). Jun 17, 2012 at 13:54
• Sure, but if the terms are tending to zero and we parenthesise the series where each parentheses contains a bounded number of terms then the parenthesised series will "behave" like the original series. Jun 17, 2012 at 13:57
• Oops. Just read that again. Never mind. Jun 17, 2012 at 13:58
• With no explicit information, it is quite pointless. If $\sum a_n$ converges conditionally, then $\sum a_{n_k}$ need not to converge, and it might even diverge. I see no parenthesis, did you forget to add some info?
– Pedro
Jun 17, 2012 at 14:03
• @Mike Since the series is infinite, the sum might be altered by conditional convergence
– Pedro
Jun 17, 2012 at 14:12

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
$$|T_n-S_n|\leq\sum_{n-M}^{n-1}|a_n|$$
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 $0$.