Check whether $\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{\left(m+n\right)^2}$ converges or NOT?

If the series were convergent, we would have

$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(m+n)^2} \geqslant\sum_{m=1}^\infty \sum_{n=1}^m \frac{1}{(m+n)^2} \geqslant \sum_{m=1}^\infty \sum_{n=1}^m \frac{1}{(2m)^2} \geqslant \sum_{m=1}^\infty \frac{1}{4m},$$

leading to a contradiction since the harmonic series on the right is divergent.

By double counting we have

$$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{\left(m+n\right)^2}=\sum_{k=2}^\infty \frac{k-1}{k^2}=\sum_{k=2}^\infty \frac{1}{k}-\sum_{k=2}^\infty \frac{1}{k^2}$$

therefore the given series diverges.

Fix $k$;

$k=m+n$; $k \ge 3$;

there are $(k-1)$ elements with $m+n=k$;

$m=k-1, n=1$; $m=k-2, n=2$;

$m=1, n=k-1;$

$\sum_n\sum_m \dfrac{1}{(m+n)^2}= \sum_{k=3}^{\infty} \dfrac{(k-1)}{k^2}$

Divergent.

From $$\frac1{(m+n)^2}=-\int_0^1 x^{m+n-1}\ln xdx$$

it follows that

$$S=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{(m+n)^2}=-\int_0^1\frac{\ln x}{x}\left(\sum_{m=1}^\infty x^m\right)\left(\sum_{n=1}^\infty x^n\right)dx=-\int_0^1\frac{x\ln x}{(1-x)^2}dx\\=-\int_0^1\ln x\left(\frac1{(1-x)^2}-\frac1{1-x}\right)dx$$

Clearly, the first integral is divergent so $S$ diverges,

$$\int_0^1\frac{\ln x}{(1-x)^2}dx=\sum_{n=1}^\infty n \int_0^1 x^{n-1}\ln xdx=-\sum_{n=1}^\infty \frac{n}{n^2}\\=-\lim_{k\to \infty}\sum_{n=1}^k \frac1n=-\lim_{k\to \infty}H_k=-\infty$$

where $H_k$ is the harmonic number.