<s>[INST] Is there any double series that cannot exchange the sum? I want to find a sequence $(u_{n,p})_{(n,p)\in\mathbb{N}^2}$ that satisfied:
$$
\sum_{n=0}^{\infty}\sum_{p=0}^{\infty}u_{n,p} ~\text{is convergent}
$$
$$
\sum_{p=0}^{\infty}\sum_{n=0}^{\infty}u_{n,p} ~\text{is convergent too}
$$
but
$$
\sum_{p=0}^{\infty}\sum_{n=0}^{\infty}u_{n,p} \neq \sum_{n=0}^{\infty}\sum_{p=0}^{\infty}u_{n,p}
$$
 [/INST] Let  $u_{mn} = 1/(m^2 - n^2)$ if $m \neq n$ and $u_{mn} =0$ if $m = n.$
Note that
$$\frac{\pi^2}{12} = \sum_{n=1}^\infty \sum_{m=1, m \neq n}^\infty \frac{1}{m^2 - n^2} \neq \sum_{m=1}^\infty \sum_{n=1, n \neq m}^\infty \frac{1}{m^2 - n^2} = - \frac{\pi^2}{12}.$$
By anti-symmetry, the double sum changes sign with an interchange of indices.
To find the sum, use
$$\begin{align}\sum_{m=1, m \neq n}^\infty \frac{1}{m^2 - n^2} &= \lim_{M \to \infty} \frac{1}{2n} \sum_{m=1, m \neq n}^M \left(\frac{1}{m-n} - \frac{1}{m+n} \right) \\ &= \lim_{M \to \infty} \frac{1}{2n} \left(\sum_{k=1}^{M-n} \frac{1}{k} -  \sum_{k=1}^{n-1} \frac{1}{k} -  \sum_{k=n+1}^{M+n} \frac{1}{k}\right)\\ &= \lim_{M \to \infty} \frac{1}{2n} \left(\frac{1}{n}-  \frac{1}{M-n+1} - \ldots - \frac{1}{M+n} \right) \\ &= \frac{1}{2n^2} \end{align}$$
</s>