$\lim_{N\to+\infty}\sum_{n=-N}^Na_n$ 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.
$\lim_{N\to+\infty}\sum_{n=-N}^Na_n$ 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.
$\displaystyle \sum_{-\infty}^{\infty} a_n$ 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.