Interchanging limit and infinite sum of a two-variable function $\lim_{x\to\infty} \sum_{n} \frac{x}{x^2 + yn^2}$

Question

Let $f(x,y) = \sum_{n} \frac{x}{x^2 + yn^2}$. Show that $g(y) = \lim_{x\rightarrow \infty} f(x,y)$ exists for all $y>0$. Find $g(y)$.

My first impression of this problem is to use the monotone convergence theorem (MCT) or the dominated convergence theorem (DCT) to interchange the limit and the sum. However, I do not know what to bound the function by. Thanks in advance!

If we convert the sum to an integral under the counting measure $\mu$ on $\mathbb{N}$, we can re-express the function as $$f(x,y) = \int_{\mathbb{N}} \frac{x}{x^2+yn^2} d\mu$$

Ideally, we want to apply the DCT by finding an integrable function $h$ such that $|f_n| = \left|\frac{x}{x^2+yn^2} \right| \leq h$ a.e. for all $n$. But, since $h$ has to be an $L^1$ function independent of $n$ under the counting measure, DCT might not be the correct method.

Also, the DCT that I learned involves interchanging $\int \lim_{n\rightarrow \infty} f_n = \lim_{n\rightarrow \infty} \int f_n$. But, this problem statement is asking for $\lim_{x\rightarrow \infty}$ and not $\lim_{n\rightarrow \infty}$.

Answer 1

We can calculate $g(y)$ directly instead of interchanging limit and infinite sum: $$\begin{aligned} g(y)&=\lim\limits_{x\rightarrow\infty}\sum\limits_{n=1}^{x}\dfrac{x}{x^{2}+yn^{2}}\\ &=\lim\limits_{x\rightarrow\infty}\dfrac{1}{x}\sum\limits_{n=1}^{x}\dfrac{x^{2}}{x^{2}+yn^{2}}\\ &=\lim\limits_{x\rightarrow\infty}\dfrac{1}{x}\sum\limits_{n=1}^{x}\dfrac{1}{1+y(\frac{n}{x})^{2}}\\ &=\int_{0}^{1}\dfrac{1}{1+yx^{2}}dx\\ &=\dfrac{1}{\sqrt{y}}\int_{0}^{1}\dfrac{1}{1+(\sqrt{y}x)^{2}}d\sqrt{y}x\\ &=\dfrac{1}{\sqrt{y}}\arctan\sqrt{y}x\big{|}_{0}^{1}=\dfrac{\arctan\sqrt{y}}{\sqrt{y}}.\\ \end{aligned}$$