4
$\begingroup$

$a_n,b_n\in\mathbb{C}$. Is it enough that both $\sum_{n\in\mathbb{Z}}a_n$ and $\sum_{n\in\mathbb{Z}}b_n$ converge?

It seems to me that this should be enough, because: $$\sum_{k\in\mathbb{Z}}(a_k+b_k)=\lim_{n\rightarrow+\infty}\sum_{k=-n}^n(a_k+b_k)=\lim_{n\rightarrow+\infty}\sum_{k=-n}^n(a_k+b_k)=\lim_{n\rightarrow+\infty}\sum_{k=-n}^na_k+\sum_{k=-n}^nb_k$$ Now it's just $$\lim x_n+y_n=\lim x_n+\lim y_n$$ which is true if the right hand side makes sense. Is my assumption correct?

  • 3
    Note that $\sum_{n \in \Bbb Z}$ and $\lim_{k \to \infty}\sum_{n = -k}^k$ are not the same thing. – 2017-01-23
  • 0
    I had no idea. Could you please elaborate further? – 2017-01-23
  • 0
    $\sum_{k \in \mathbb Z}$ is not necessarily the same thing as $\lim_{n \to \infty} \sum_{k=-n}^{n}$. Actually $\sum_{k \in \mathbb Z} = \lim_{m,n \to \infty} \sum_{k=-n}^{m}$, and this subtle point causes some trouble with your argument. – 2017-01-23
  • 1
    Take, for instance, the sequence $a_n = 1/n$ (and $a_0 = 0$ or something). Then $\lim_{k \to \infty}\sum_{n = -k}^k a_n = 0$ while $\sum_{n \in \Bbb Z}a_n$ doesn't exist. Intuitively, $\sum_{n \in \Bbb Z}$ exists if the _ratio_ between how the lower and upper limit go to their respective infinities doesn't affect anything.. – 2017-01-23
  • 0
    Ok, but if $\sum_{n\in\mathbb{Z}}$ exists, then it's equal to $\lim_{n\rightarrow\infty}\sum_{k=-n}^n$ isn't it? The equality should still hold then, right? – 2017-01-23
  • 0
    @Tom83B Yes, and no. My point is just that it doesn't go the other way around. Your argument has to take into account that for $\sum_{n \in \Bbb Z} (a_n+b_n)$ to exist, the limit has to exist no matter how the lower and upper bound goes to infinity, and you have to make full use of the fact that it _is_ true for $\sum_{n \in \Bbb Z}a_n$ and $\sum_{n \in \Bbb Z}b_n$ – 2017-01-23

0 Answers 0