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?