Why does $\sum_{i,j}^\infty \frac{1}{i(i^2+j^2)}$ converge? According to Mathematica the series $\sum_{i,j=1}^\infty \frac{1}{i(i^2+j^2)}$ converges. Is there a simple argument to obtain the convergence? Maybe a dominating convergent series?
 A: By the integral test
$$\sum_{i=1}^\infty \frac{1}{i^3+ij^2}$$
is convergent and
$$\sum_{i=1}^\infty \frac{1}{i^3+ij^2} \leq \frac{1}{1+j^2}+ \int_{x=1}^\infty \frac{1}{x^3+xj^2} dx = \frac{1}{1+j^2}+ \int_{x=1}^\infty \frac{x}{x^4+x^2j^2} dx  \\
=\frac{1}{1+j^2}+ \frac{1}{2}\int_{u=1}^\infty \frac{1}{u(u+j^2)} du \\ =\frac{1}{1+j^2}+ \frac{1}{2}\lim_R \to \infty \int_1^R \frac{1}{j^2} \left( \frac{1}{u}-\frac{1}{u+j^2} \right) du \\=\frac{1}{1+j^2}+ \frac{1}{2j^2} \lim_R\left( \ln(u)-\ln(u+j^2) \right)_1^\infty=\frac{1}{1+j^2}+ \frac{\ln(1+j^2)}{2j^2}$$
Next, 
$$\sum_{j=1}^\infty \frac{\ln(1+j^2)}{2j^2} < \infty$$
by (Limit) comparison test (compare to $\sum \frac{1}{j^{\frac{3}{2}}}$)
and $\sum_{j=1}^\infty \frac{1}{1+j^2}$ is convergent.
A: Divide your sum into $i \leq j$ and $i > j$ parts:
$$\sum_{i \leq j} {1 \over i(i^2 + j^2)} + \sum_{i > j} {1 \over i(i^2 + j^2)}$$
In the first sum use that $i^2 + j^2 > j^2$ and in the second sum use that $i^2 + j^2 > i^2$ to get a bound of
$$\sum_{\{(i,j):\, i \leq j\}} {1 \over ij^2} + \sum_{\{(i,j):\, i > j\}} {1 \over i^3}$$
In the first sum, the $i$ summation is bounded by $C\ln j$, so since the sum of ${\ln j \over j^2}$ is finite, the first sum converges. In the second sum, the sum in $j$ gives you a factor of $i$, so since the sum in $i$ of ${1 \over i^2}$ is finite the second sum converges as well.
