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?


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.