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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.

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  • $\begingroup$ Any reason for omitting the $i=j$ part? $\endgroup$ – Suvrit Jul 31 '13 at 5:05
  • $\begingroup$ From a real problem. But probably the same with $i=j$ part. $\endgroup$ – Uchiha Jul 31 '13 at 11:03
  • $\begingroup$ Just an aside: Near the end, the inequality $i^\alpha + j^\beta \geq 2 (ij)^{(\alpha+\beta)/2}$ is false in general. Perhaps you intended the right-hand side to be $2 i^{\alpha/2} j^{\beta/2}$? $\endgroup$ – cardinal Aug 1 '13 at 2:27
  • $\begingroup$ That's true. Thanks for pointing out. I've made the change. $\endgroup$ – Uchiha Aug 1 '13 at 11:46
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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.

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  • $\begingroup$ The arguments seem all correct. Thanks, Matt. Now the question becomes if we have $S(n)\sim c n^{1-2\beta+\beta/\alpha}$, or in a weaker statement, if $S(n)$ is regularly varying with that exponent. $\endgroup$ – Uchiha Aug 1 '13 at 21:33

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