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?

  • $\begingroup$ Do you know how to approximate an integral by a series and vice versa? e.g., see the pictures here $\endgroup$
    – angryavian
    Commented Oct 13, 2021 at 16:56
  • $\begingroup$ I understand how sums and integrals can be approximated by each other. I can vaguely see how doing so would lead to asymptotic equality, but I still feel uneasy. Could it be the case that the error term in such an approximation mucks up asymptotic equality? $\endgroup$
    – eeen
    Commented Oct 13, 2021 at 17:14

2 Answers 2


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 $2\epsilon \to 0$.

  • $\begingroup$ Thanks! It seems a similar argument follows for the asymptotic equality between a double sum and a double integral. $\endgroup$
    – eeen
    Commented Oct 13, 2021 at 18:13

I've found an answer that I'm more satisfied with than the other one. It generalizes the process.

Claim. For each $\varepsilon > 0$, suppose we have a function $f_\epsilon : [1, \infty) \to (0, \infty)$ that is both continuous and nonincreasing. Set

$$S(\varepsilon) = \sum_{n=1}^\infty f_\varepsilon(n) \qquad \text{and} \qquad T(\varepsilon) = \int_1^\infty f_\varepsilon(x) \ dx.$$

Assume $S(\varepsilon) \to \infty$ and $T(\varepsilon) \to \infty$ as $\varepsilon \to 0$. Then, if $f_\varepsilon(1) = o(S(\varepsilon))$ as $\varepsilon \to 0$, then $S(\varepsilon) \sim T(\varepsilon)$.

Proof. Since $f_\varepsilon$ is nonincreasing, we may write

$$\sum_{n=2}^\infty f_\varepsilon(n) \leq \int_1^\infty f_\varepsilon(x) \ dx \leq \sum_{n=1}^\infty f_\varepsilon(n),$$

which is notationally equivalent to

$$S(\varepsilon) - f_\varepsilon(1) \leq T(\varepsilon) \leq S(\varepsilon).$$

Divide out by $S(\varepsilon) > 0$ to obtain

$$1 - \frac{f_\varepsilon(1)}{S(\varepsilon)} \leq \frac{T(\varepsilon)}{S(\varepsilon)} \leq 1.$$

Finally, $f_\varepsilon(1) / S(\varepsilon) \to 0$ as $\varepsilon \to 0$. So, the above shows that $T(\varepsilon)/S(\varepsilon) \to 1$ as $\varepsilon \to 0$, and hence $T(\varepsilon) \sim S(\varepsilon)$.

In the example of the double sum in question, we can write

$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac1{n^{\frac12 + \varepsilon}} \frac1{m^{\frac12 + \varepsilon}} \frac1{m+n} \sim \int_1^\infty\int_1^\infty \frac1{x^{\frac12 + \varepsilon}} \frac1{y^{\frac12 + \varepsilon}} \frac1{x+y} \ dx \ dy$$

since the function $F(m,n) = 1/n^{1/2 + \epsilon} \cdot 1/m^{1/2 + \epsilon} \cdot 1/(m+n)$ satisfies that

  • $F$ is decreasing in $m$ and $n$, separately,
  • $F$ is continuous for all $m,n \geq 1$,
  • Both the sum and integral over $F$ go to infinity as $\varepsilon \to 0$ (compare with harmonic series),
  • $F(1,n)$ and $F(m,1)$ both stay finite as $\varepsilon \to 0$.

I don't know if these conditions are necessary to give asymptotic equality between a sum and a corresponding integral, but it certainly works in the cases used in the book.


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