Grouping terms of a series with the same sign Let $\displaystyle \sum_{n=1}^{\infty} a_n$ be a series with infinitely many positive terms and infinitely many negative terms, 
and
let $\displaystyle \sum_{n=1}^{\infty} b_n$ be the series obtained by grouping together each string of consecutive terms with the same sign.
For example, let $\displaystyle \sum_{n=1}^{\infty} a_n=1+1-2-2+3+3+3+3-4-4-4+5-6-6+\cdots$,
in which case $\displaystyle \sum_{n=1}^{\infty} b_n=(1+1)-(2+2)+(3+3+3+3)-(4+4+4)+(5)-(6+6)+\cdots$
If $\displaystyle \sum_{n=1}^{\infty} b_n$ converges, does it follow that $\displaystyle \sum_{n=1}^{\infty} a_n$  converges to the same sum?
(See Property of series (grouping) for a similar question.)
 A: As long as the terms are not rearranged, the partial sums of the series $\sum b_n$ is just a sequence of some of the partial sums of $\sum a_n$ in which one stops after each sign change. This in itself (as user vadim123 pointed out in his comment) doesn't imply that $\sum a_n$ converges follows from $\sum b_n$ converging. However the partial sums of $a_n$ in this setup lie between two adjacent partial sums of the $b_n$ so that we do get $\sum a_n$ converges from $\sum b_n$ converging. To illustrate:
$a_n=p_1,p_2,p_3,-p_4,-p_5,p_6,..$ where each $p_k$ is positive. Then $b_1=p_1+p_2+p_3$ and $b_2=-p_4-p_5.$ The fourth partial sum of $\sum a_n$ is then $b_1-p_4$ which lies between $b_1$ and $b_2=b_1-p_4-p_5.$
So IF the partial sums of the $b_k$ converge, so do the partial sums of the $a_k$ using the squeeze theorem.
A: Let $(S_n)$ and $(T_n)$ be the sequences of partial sums for $\displaystyle\sum_{n=1}^{\infty}a_n$ and $\displaystyle\sum_{n=1}^{\infty}b_n$, respectively.
Since $\displaystyle\sum_{n=1}^{\infty}b_n$ converges, $\displaystyle\lim_{n\to\infty}T_n=T$ for some number $T$; 
so for any $\epsilon>0$, there is an $N$ such that $n\ge N\implies |T_n-T|<\epsilon$.
Then $T_N=S_M$ for some $M$; 
and if $m\ge M$, then $\;T_n\le S_m<T_{n+1}\;\;$ or $\;\;T_n\ge S_m>T_{n+1}\;$ for some $n\ge N$.
Therefore $m\ge M\implies |S_m-T|\le \max\{|T_n-T|, |T_{n+1}-T|\}<\epsilon$; 
so $\displaystyle\lim_{n\to\infty}S_n=T$ and therefore $\displaystyle\sum_{n=1}^{\infty}a_n$ converges.
