I am trying to find a counter-example to the statement:

$$\lim_{N \to \infty} \sum_{k=-N}^{N}\alpha_k \:\:\: \text{exists} \implies \sum_{k=-\infty}^{\infty}\alpha_k \:\:\: \text{converges.}$$

I took a very simple sequence as a counter-example, $(\alpha)_k = k$, $k \in \mathbb{Z}.$ Now I have for the limit of partial sums

$$\lim_{N \to \infty} \sum_{k=-N}^{N}\alpha_k = \lim_{N \to \infty} \sum_{k=-N}^{N}k $$

And if I break the sum into two I get

$$\lim_{N \to \infty} \sum_{k=-N}^{N}k = \lim_{N \to \infty} \Big[ \sum_{k=0}^{N}k + \sum_{k=1}^{N}-k \Big] = 0$$

Now the limit of partial sums converges to $0$, however sums $\sum_{k=1}^{\infty}-k$ and $\sum_{k=0}^{\infty}k$ diverge to $-\infty$ and $\infty$ respectively. I think this means that "by definition" $\sum_{k=-\infty}^{\infty}k$ must also diverge.

I have two questions. Firstly, does this counter-example work? And secondly I don't understand why having $\sum_{k=1}^{\infty}-\alpha_k$ and $\sum_{k=0}^{\infty}\alpha_k$ divergent necessarily means that $\sum_{k=-\infty}^{\infty}\alpha_k$ diverges.

If $\sum_{k=1}^{\infty}-\alpha_k + \sum_{k=0}^{\infty}\alpha_k = 0$ then why not $\sum_{k=-\infty}^{\infty}\alpha_k = 0$?

  • $\begingroup$ At some point you have to go back to the definitions themselves. The simple fact is that $\sum_{k=-\infty}^\infty a_k =L$ means for most of us that, for every $\epsilon>0$, there are positive integers $M$ and $N$ so that $|\sum_{k=m}^n a_k -L|<\epsilon$ for all $n \geq N$ and for all $m \leq -M$. You are trying to rely on your intuition as to how this works. Work with the definition often enough and then your intuition kicks in and you can do these things more easily. But you are definitely on the right track here. $\endgroup$ – B. S. Thomson Oct 23 '15 at 6:25
  • $\begingroup$ What's:$$\lim_{N\to\infty}\sum_{k=-N}^{N+1}k$$For the sum $\sum_{-\infty}^\infty$ to exist, it can't matter how the two indices approach infinity. $\endgroup$ – Akiva Weinberger Oct 23 '15 at 15:14


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