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$\lim_{N\to+\infty}\sum_{n=-N}^Na_n$ exists but $\sum_{-\infty}^{+\infty}a_n$ does not converge, where $a_n$ are complex constants.

If let $a_n=i$, it looks fine but way too easy. I wonder if there is better solutions.

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It is incorrect to say "$\sum_{-\infty}^{+\infty}a_n$ does not converge". You must say "$\sum_{-\infty}^{+\infty}a_n$ doesn't exist". –  user17762 Oct 30 '12 at 23:43

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up vote 3 down vote accepted

$\displaystyle \sum_{-\infty}^{\infty} a_n$ denotes $$\displaystyle \lim_{\substack{N_1 \to \infty\\ N_2 \to \infty}}\sum_{-N_1}^{N_2} a_n$$

If you consider $$a_n = \begin{cases} 0 & \text{if }n=0\\ 1/n & \text{if }n \in \mathbb{Z} \backslash \{0\} \end{cases}$$ we get that $\displaystyle \sum_{-N}^{N} a_n = 0$ and hence $$\lim_{N \to \infty} \sum_{-N}^{N} a_n = 0$$ but $$\displaystyle \sum_{-\infty}^{\infty} a_n$$ doesn't exist.

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Thank you very much, Marvis. Your explanation makes me much clearer. –  Sam Oct 30 '12 at 23:47

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