Calculate the Limit of Double Sum

Question

Compute \begin{equation} L=\lim _{n \rightarrow \infty}\frac{1}{n} \sum_{a=1}^n \sum_{b=1}^n \frac{a}{a^2+b^2 }. \end{equation} My attempt: Define \begin{equation} f(n,m)= \frac{1}{n} \sum_{a=1}^n \frac{1}{m}\sum_{b=1}^m \frac{\frac{a}{n}}{(\frac{a}{n})^2+(\frac{b}{m})^2 } = \frac{1}{m}\sum_{b=1}^m \frac{1}{n} \sum_{a=1}^n \frac{\frac{a}{n}}{(\frac{a}{n})^2+(\frac{b}{m})^2 } \end{equation} Now for any $\epsilon >0$ there exists some $B>0$ such that for any $n,m \in \mathbb{N}$ and $n,m\geq B$: \begin{equation} |f(n,m)-f(n,n)|<\epsilon \end{equation} Thus, we have \begin{equation} L=\lim _{n \rightarrow \infty} f(n,n)=\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty} f(n,m)= \lim _{m \rightarrow \infty} \frac{1}{m}\sum_{b=1}^m \int_{0}^{1}\frac{x}{x^2+(\frac{b}{m})^2}dx=\\ \frac{1}{2}\lim _{m \rightarrow \infty} \frac{1}{m}\sum_{b=1}^m \ln\frac{1+(\frac{b}{m})^2}{(\frac{b}{m})^2}=\frac{1}{2}\int_{0}^{1}\ln\frac{1+x^2}{x^2}dx=\frac{2\ln2 +\pi}{4} \end{equation}

Answer 1

While this approach arrives at the correct value for the limit it is not easily justified. At first glance,

$$f(n,n) = \frac{1}{n}\sum_{a=1}^n\sum_{b=1}^n\frac{a}{a^2 + b^2} = \frac{1}{n^2}\sum_{a=1}^n\sum_{b=1}^n\frac{a/n}{(a/n)^2 + (b/n)^2} $$

is a Riemann sum for the integral of $g:(x,y) \mapsto \frac{x}{x^2+ y^2}$ over $[0,1]\times[0,1]$. However, the integrand is unbounded around $(0,0)$ and is not Riemann integrable. Instead it must be considered as an improper integral which can be evaluated as an iterated integral $$I = \int_0^1 \left(\int_0^1 \frac{x}{x^2 + y^2} \, dy\right )dx = \frac{2 \log 2 + \pi}{4}$$

Note that the inner integral is properly Riemann and we can write

$$I = \int_0^1 \lim_{m \to \infty} \left(\frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 + (b/m)^2} \right)\, dx$$

At this point it would be nice to switch the integral and the limit and apply a second Riemann sum to obtain

$$I = \lim_{m \to \infty} \frac{1}{m}\sum_{b=1}^m \int_0^1\frac{x}{x^2 + (b/m)^2} \, dx = \lim_{m \to \infty} \left(\frac{1}{m}\lim_{n \to \infty} \frac{1}{n}\sum_{a=1}^n\sum_{b=1}^m \frac{a/n}{(a/n)^2 + (b/m)^2}\right)$$

However justifying that switch is not obvious. For example, we see immediately that

$$\left|\frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 + (b/m)^2} \right| \leqslant \frac{1}{m}\sum_{b=1}^m \frac{x}{x^2 } = \frac{1}{x},$$

but the dominating function $x \mapsto 1/x$ is not integrable on $[0,1]$.

Furthermore, there remains finding the justification for $\lim_{n\to \infty} f(n,n) = \lim_{m\to \infty} \lim_{n\to \infty} f(n,m)$.

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A easily justified approach is to apply the Stolz-Cesaro theorem to get

$$\lim_{n \to \infty} f(n,n) = \lim_{n \to \infty} \frac{\sum_{a=1}^{n+1}\sum_{b=1}^{n+1}\frac{a}{a^2 + b^2} - \sum_{a=1}^n\sum_{b=1}^n\frac{a}{a^2 + b^2} }{n+1 - n} \\ = \lim_{n \to \infty} \left(\frac{n+1}{(n+1)^2 + (n+1)^2}+ \sum_{a=1}^n \frac{a}{a^2 + (n+1)^2} + \sum_{b=1}^n \frac{n+1}{(n+1)^2 + b^2} \right) \\ = \lim_{n\to \infty} \sum_{a=1}^{n} \frac{a + n+1}{a^2 + (n+1)^2} = \lim_{n\to \infty} \frac{1}{n+1}\sum_{a=1}^{n} \frac{1 + \frac{a}{n+1}}{1 + \left(\frac{a}{n+1}\right)^2}\\ = \int_0^1 \frac{1+x}{1+x^2} \, dx = \frac{2 \log 2 + \pi}{4}$$