Let $f(x,y)=\sum _{n}\frac{x}{{x}^{2}+y{n}^{2}}$. Show that $g(y)=\underset{x\to \mathrm{\infty}}{lim}f(x,y)$ 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}+y{n}^{2}}d\mu $

Ideally, we want to apply the DCT by finding an integrable function $h$ such that $|{f}_{n}|=\left|\frac{x}{{x}^{2}+y{n}^{2}}\right|\le h$ 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 $\int \underset{n\to \mathrm{\infty}}{lim}{f}_{n}=\underset{n\to \mathrm{\infty}}{lim}\int {f}_{n}$. But, this problem statement is asking for $\underset{x\to \mathrm{\infty}}{lim}$ and not $\underset{n\to \mathrm{\infty}}{lim}$.

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}+y{n}^{2}}d\mu $

Ideally, we want to apply the DCT by finding an integrable function $h$ such that $|{f}_{n}|=\left|\frac{x}{{x}^{2}+y{n}^{2}}\right|\le h$ 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 $\int \underset{n\to \mathrm{\infty}}{lim}{f}_{n}=\underset{n\to \mathrm{\infty}}{lim}\int {f}_{n}$. But, this problem statement is asking for $\underset{x\to \mathrm{\infty}}{lim}$ and not $\underset{n\to \mathrm{\infty}}{lim}$.