Let $1<\alpha<\beta<3/2$. Set
$$
S(n)= \sum\_{i,j>0} [i^\alpha+j^\beta]^{-1}[(i+n)^\alpha+(j+n)^\beta]^{-1}.
$$
One can check that $S(n)$ is finite. My question is when $n\rightarrow \infty$, how does $S(n)$ behave asymptotically, e.g., if it is asymptotically a power function? If yes, what is the exponent? 


Remarks:


When $\alpha=\beta$, this problem can be resolved using an integral approximation argument (rewriting the sum as a double integral by replacing $\frac{i}{n}$ with $\frac{[nx]+1}{n}$, $\frac{j}{n}$ with $\frac{[ny]+1}{n}$ and letting $n\rightarrow\infty$ through the Dominated Convergence Theorem) which yields $S(n)\sim c n^{2-2\alpha}$ for some $c>0$. But when $\alpha<\beta$, the similar argument seems difficult to apply due to the non-homogeneity of the function $g(x,y)=(x^{\alpha} +y^{\beta})^{-1}$.


It seems that if we do have $S(n)\sim cn^{2-2\gamma}$ for some $\gamma$, then $\alpha\le \gamma\le \beta$. Furthermore, by Jensen's inequality, we have for any $0<\rho<1$, $i^\alpha+j^\beta\ge c i^{\alpha\rho}j^{\beta(1-\rho)}$ (now $g(x,y)= x^{-\rho\alpha}y^{-(1-\rho)\beta}$ is homogeneous, and an integral approximation argument applies provided $\alpha\rho\in (1/2,3/4)$, $\beta(1-\rho)\in (1/2,3/4)$), we should have 
$
\gamma\ge\rho\alpha+(1-\rho)\beta. 
$
By taking $\rho$ close to $1/(2\alpha)$, we expect that $\gamma\ge \beta+(\alpha-\beta)/(2\alpha)$.


**Update**: Matt shows below that $cn^{2-2\gamma}\le S(n)\le C n^{2-2\gamma}$, where $$\gamma=\beta+\frac{\alpha-\beta}{2\alpha}=\rho\alpha+(1-\rho)\beta\in (\alpha,\beta),$$ with $\rho=\frac{1}{2\alpha}$. Now the problem becomes whether one can show that $S(n)\sim cn^{2-2\gamma}$ where $\gamma$ is given as above. 



Certain asymptotics involving double infinite sum
I can show without too much work that there exist absolute constants $c, C > 0$ so that $$c \leq \frac{S(n)}{n^{1-2\beta + \frac{\beta}{\alpha}}} \leq C.$$ This at least shows what the exponent must be if there is an asymptotic formula for $S(n)$. When $\alpha = \beta$ this gives $S(n) \asymp n^{2-2\alpha}$, which uncovers a typo in the remarks in the question. Here and below the notation $f(x) \asymp g(x)$ for positive functions $f$ and $g$ means there exist $c, C > 0$ so that $c < \frac{f(x)}{g(x)} < C$.


Proof. First note that $$(i + n)^{\alpha} + (j + n)^{\beta} \asymp i^{\alpha} + j^{\beta} + n^{\beta}.$$ The terms with $i^{\alpha} \leq j^{\beta}$ and $j \leq n$ contribute to $S(n)$ an amount
$$\asymp \sum\_{j \leq n} \sum\_{i \leq j^{\beta/\alpha}} \frac{1}{i^\alpha + j^{\beta}} \frac{1}{i^{\alpha} + j^{\beta} + n^{\beta}}$$ which is
$$\asymp \sum\_{j \leq n} \frac{1}{j^{\beta}} \frac{1}{j^{\beta} + n^{\beta}} \sum\_{i \leq j^{\beta/\alpha}} 1 \asymp n^{-\beta} \sum\_{j \leq n} \frac{1}{j^{\beta -\frac{\beta}{\alpha}}}.$$ Using $0 < \beta - \frac{\beta}{\alpha} < 1$, it is not hard to check that this is $\asymp n^{1-2\beta + \frac{\beta}{\alpha}}$. Similarly, the terms with $i^{\alpha} \leq j^{\beta}$ and $j > n$ satisfy the same type of estimates, as do the terms with $i^{\alpha} \geq j^{\beta}$ where in this case one should execute the $j$-sum first and divide the $i$ sum into two pieces depending on if $i^{\alpha} \leq n^{\beta}$ or not.


