Check whether $\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{\left(m+n\right)^2}$ converges or NOT? Check whether $$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{\left(m+n\right)^2}$$ converges or NOT?
My Try:- $\sum _{m=1}^{\infty }\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(m+n\right)^2}=\lim_{j\to \infty}\sum _{m=1}^{j }\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(m+n\right)^2}=\lim_{j\to \infty}(\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(1+n\right)^2}+ \lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(2+n\right)^2}+ \lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(3+n\right)^2}+...\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(j+n\right)^2})\ge \lim_{j\to \infty}(j\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(j+n\right)^2})$
How do I complete the conclusion?
 A: 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.
A: 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.

A: 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.
A: 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.
