Let $a_n \in \mathbb{C}$ and consider $\sum a_n$ and grouping as $\sum (a_n + a_{n+1})$.
Under what assumptions we can claim absolute convergence of grouped sum implies convergence of the original sum?

Here is the related post, where accepted answer shows that grouped sum is absolutely convergent.

My concern is, $\sum (-1)^n$ is not convergent where if we group successive terms it is absolutely convergent.


$\sum |a_n| <\infty$ implies $|a_1-a_2|+|a_3-a_4|+\cdots <\infty$ but the converse is false as seen by taking $a_1=a_2=1, a_3=a_4=\frac 1 2,a_5=a_6=\frac 1 3,\cdots$.

  • $\begingroup$ Is it then true that convergence of $1^{-z} - 2^{-z} + 3^{-z} - \dots$ cannot be showed by arguing $(1-^{-z} - 2^{-z}) + (3^{-z} - 4^{-z}) + \dots$ is absolutely convergent. (As in the first way of accepted answer of linked question) $\endgroup$ – Jo' May 6 at 0:03
  • $\begingroup$ You are right. Proving (absolute0 convergence after grouping does nor tell you anything about convergence of the original series. $\endgroup$ – Kavi Rama Murthy May 6 at 0:06

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.