Show that limit function exists and evaluate the limit function

Question

Let $$f(x,y) = \sum_{n=1}^\infty \frac{x}{x^2+yn^2}, y > 0$$

(i) Show that for each $y$, the limit $$g(y):=\lim_{x\to\infty} f(x,y)$$ exists. Evaluate the limit function $g$.

I found this problem on Wisconsin’s analysis qual study guide. I have tried to find the maximum value of $\frac{x}{x^2+yn^2}$ for a given $y$ and $n$, and I found that it is equal $\frac{1}{2\sqrt{y}n}$, when $x=\sqrt{y}n$. Beyond that, I haven’t been able to do much. There is a second part to this question:

(ii) Determine if $f(x,y)$ converges to $g(y)$ uniformly for $y\in(0,\infty)$ as $x\to \infty$.

Thank you for your help!

Answer 1

Part i) For fixed $x,y$ the function $t\to \frac{x}{x^2+yt^2}$ is decreasing on $[0,\infty).$ Thus

$$\tag 1 \int_1^\infty \frac{x}{x^2+yt^2}\, dt \le \sum_{n=1}^{\infty}\frac{x}{x^2+yn^2} \le \int_0^\infty \frac{x}{x^2+yt^2}\, dt.$$

Let $t=xu/\sqrt y$ in the integral on the right of $(1).$ The integral becomes

$$\frac{1}{\sqrt y}\int_0^\infty \frac{1}{1+u^2}\, du = \frac{\pi}{2\sqrt y}.$$

It's not hard to see the integral on the left of $(1)$ converges to the integral on the right of $(1)$ as $x\to \infty.$ Thus the limit of the sum in $(1)$ is $\dfrac{\pi}{2\sqrt y}.$

Answer 2

Hoping that this is not too advanced.

Consider the partial sum

$$S_p=\sum_{n=1}^p \frac{x}{x^2+yn^2}=\frac x y \sum_{n=1}^p \frac{1}{k^2+n^2}\qquad \text{with}\qquad k^2=\frac {x^2} y$$ Uing partial fraction decomposition $$\frac{1}{k^2+n^2}=\frac{1}{(n+ik)(n-ik)}=\frac i{2k}\Bigg[\frac{1}{n+ik}- \frac{1}{n-ik}\Bigg]$$ $$\sum_{n=1}^p \frac{1}{n+ik}=H_{p+i k}-H_{i k}\qquad \text{and}\qquad \sum_{n=1}^p \frac{1}{n-ik}=H_{p-i k}-H_{-i k}$$ Using the asymptotics of generalized harmonic numbers $$\sum_{n=1}^p \Bigg[\frac{1}{n+ik}- \frac{1}{n-ik}\Bigg]=\left(H_{-i k}-H_{i k}\right)+\frac{2 i k}{p}+O\left(\frac{1}{p^2}\right)$$ but $$H_{-i k}-H_{i k}=\frac{i}{k}-i \pi \coth (\pi k)$$ which makes $$S_p=\frac x y \Bigg[\frac{\pi k \coth (\pi k)-1}{2 k^2}-\frac{1}{p}+O\left(\frac{1}{p^2}\right) \Bigg]$$ Replacing $k$ by its value $$S_\infty=\frac{\pi |x| \coth \left(\frac{\pi |x|}{\sqrt{y}}\right)}{2 x \sqrt{y}}-\frac{1}{2 x}$$ If $x \to \infty$ this will give $\frac{\pi }{2 \sqrt{y}}$