Determine the largest real number in [0,2] $\beta$ a double series converges.

Question

I wish to determine the largest real number $\beta \in [0,2]$ for which the following converges: \begin{align} \sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \frac{(i^2+j^2)^{\beta-1}}{(ij)^3} \end{align} now, my first attempt is either to try to say, ok, $\beta=2$ and then see that $\beta$ is to large... but I need to either show that it really is $\beta-\epsilon$ for where $\epsilon$ is arbitrarly small or determine the exact value. So, my second attempt is to use bionomial expansions, where $()_k$ is the falling factorial:

\begin{align} \sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \frac{1}{i^3j^3} \sum_{k=0}^{\infty} \frac{(\beta-1)_k}{k!}(i^2)^{\beta-1-k}(j^2)^k \end{align} and then try to move stuff in...but there i get stuck. The third attempt is to see use Cauchy's condensation criterion, but no dice. any suggestions?

Answer 1

Hints:

1. Use $$(i^2+j^2)^{\beta-1} \geq i^{2(\beta-1)}$$ to show that $$\sum_{i=1}^{\infty} \frac{(i^2+j^2)^{\beta-1}}{(ij)^3} = \infty \quad \text{for $\beta=2$}.$$ Hence, $$\sum_{j=1}^{\infty} \sum_{i=1}^{\infty} \frac{(i^2+j^2)^{\beta-1}}{(ij)^3} = \infty \quad \text{for $\beta=2$}.$$
2. Prove (or recall) that $(x+y)^p \leq x^p + y^p$ for any $x,y>0$ and $p \in (0,1)$.
3. Show that $$(i^2+j^2)^{\beta-1} \leq \begin{cases} i^{2(\beta-1)} + j^{2(\beta-1)}, & \beta \in (1,2) \\ 1, & \beta \in [0,1] \end{cases}$$ for any $i,j \geq 1$.
4. Use Step 3 to prove that $$\sum_{j=1}^{\infty} \sum_{i=1}^{\infty} \frac{(i^2+j^2)^{\beta-1}}{(ij)^3} < \infty \quad \text{for $\beta \in (0,2)$}.$$