I'm currently reading through Chapter 10 of the book The Cauchy-Schwarz Master Class.
I'm stuck on this step that they use to prove what they call the Double Sum Lemma:
Double Sum Lemma
$$ \sum_{m=1}^\infty \sum_{n=1}^\infty \frac1{n^{\frac12 + \epsilon}} \frac1{m^{\frac12 + \epsilon}} \frac1{m+n} \sim \frac\pi{2\epsilon} \qquad \text{as } \epsilon \to 0.$$
For the proof, we first note that integral comparisons tell us that it suffices to show
$$I(\epsilon) = \int_1^\infty\int_1^\infty \frac1{x^{\frac12 + \epsilon}} \frac1{y^{\frac12 + \epsilon}} \frac1{x+y} \ dx \ dy \sim \frac\pi{2\epsilon} \qquad \text{as $\epsilon \to 0$}.$$
I'm assuming that what they mean is that, as $\epsilon \to 0$,
$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac1{n^{\frac12 + \epsilon}} \frac1{m^{\frac12 + \epsilon}} \frac1{m+n} \sim \int_1^\infty\int_1^\infty \frac1{x^{\frac12 + \epsilon}} \frac1{y^{\frac12 + \epsilon}} \frac1{x+y} \ dx \ dy,$$
and that, more generally, for "certain" $f_\epsilon \colon [1, \infty) \to (0, \infty)$, as $\epsilon \to 0$,
$$\sum_{n=1}^\infty f_\epsilon(n) \sim \int_1^\infty f_\epsilon(x) \ dx.$$
I kinda chose $f_\epsilon$ arbitrarily because I don't really know what's going on here. An earlier result they also use is that, as $\epsilon \to 0$,
$$\sum_{n=1}^\infty \frac1{n^{1 + 2\epsilon}} \sim \int_1^\infty \frac1{x^{1+2\epsilon}} \ dx,$$
which I'm also kind of confused on. I only really know asymptotics to the point that $f(x) \sim g(x)$ as $x \to 0$ means to say that
$$\lim_{x \to 0} \frac{f(x)}{g(x)} = 1.$$
Can someone explain to me in more detail what's going on here? And also, when do you have asymptotic equality between the sum of a function and the integral of the same function?
Best Answer
To show $\sum_{n \ge 1} n^{-(1+2\epsilon)} \sim \int_1^\infty x^{-(1+2\epsilon)} \, dx$ it suffices to show $$\frac{\int_1^\infty (\lfloor x \rfloor^{-(1+2\epsilon)} - x^{-(1+2\epsilon)}) \, dx}{\int_1^\infty x^{-(1+2\epsilon)} \, dx} \to 0$$
The numerator is bounded by $\le \sum_{n \ge 1} (n^{-(1+2\epsilon)} - (n+1)^{-(1+2\epsilon)}) = 1$. The denominator is $\frac{1}{2\epsilon}$. So the entire expression is bounded by $2\epsilon \to 0$.