2
$\begingroup$

Lets say I have a double infinite sum $\sum\limits_{n=0}^\infty a_n\sum\limits_{k=n}^\infty b_k$ If I know that the sum is absolutely convergent and that $\sum\limits_{k=n}^\infty b_k$ is absolutely convergent, what kind of rearrangements are valid? For example I would like expand the series and collect certain terms into meaningful groups, is that a "legal" operation provided all the series I work with are absolutely convergent?

$\endgroup$
3
$\begingroup$

If the series $\sum_{n=1}^\infty |a_n| \sum_{k=n}^\infty |b_k| $ converges, then you can do whatever you want with $a_n b_k$; any rearrangement will converge to the same sum. More generally: if $I$ is an index set and $\sum_{i\in I} |c_i|<\infty$, then any rearrangement of $c_i$ converges to the same sum. This is because for every $\epsilon>0$ there is a finite set $S\subset I$ such that $\sum_{i\in I\setminus S}|c_i|<\epsilon $; any method of summation will use up $S$ at some point, and after that the sum is guaranteed to be within $\epsilon$ of $\sum_{i\in S} c_i$.

But if you only know that $$\sum |b_k|\tag{1}$$ and $$\sum_{n=1}^\infty \left|a_n \sum_{k=n}^\infty b_k \right|\tag{2} $$ converge, then rearrangements can go wrong. For example, take the series $\sum b_k $ to be $\frac{1}{2} -\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+\frac{1}{8}-\frac{1}{8}+\dots$ then every other sum $\sum_{k\ge n} b_k$ is zero, and the corresponding $a_n$ could be arbitrarily large without disturbing the convergence of (2). In this situation you could even have infinitely many terms $a_nb_k$ that are greater than $1$ in absolute value; clearly this series can't be rearranged at will.

$\endgroup$

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.