 # Let f ( x , y ) = <munder> &#x2211;<!-- ∑ --> <mrow class="MJX-Te babyagelesszj 2022-07-01 Answered
Let $f\left(x,y\right)=\sum _{n}\frac{x}{{x}^{2}+y{n}^{2}}$. Show that $g\left(y\right)=\underset{x\to \mathrm{\infty }}{lim}f\left(x,y\right)$ exists for all $y>0$. Find $g\left(y\right)$.
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
$f\left(x,y\right)={\int }_{\mathbb{N}}\frac{x}{{x}^{2}+y{n}^{2}}d\mu$
Ideally, we want to apply the DCT by finding an integrable function $h$ such that $|{f}_{n}|=|\frac{x}{{x}^{2}+y{n}^{2}}|\le h$ 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 $\underset{n\to \mathrm{\infty }}{lim}$.
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We can calculate $g\left(y\right)$ directly instead of interchanging limit and infinite sum:
$\begin{array}{rl}g\left(y\right)& =\underset{x\to \mathrm{\infty }}{lim}\sum _{n=1}^{x}\frac{x}{{x}^{2}+y{n}^{2}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}\sum _{n=1}^{x}\frac{{x}^{2}}{{x}^{2}+y{n}^{2}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}\sum _{n=1}^{x}\frac{1}{1+y\left(\frac{n}{x}{\right)}^{2}}\\ & ={\int }_{0}^{1}\frac{1}{1+y{x}^{2}}dx\\ & =\frac{1}{\sqrt{y}}{\int }_{0}^{1}\frac{1}{1+\left(\sqrt{y}x{\right)}^{2}}d\sqrt{y}x\\ & =\frac{1}{\sqrt{y}}\mathrm{arctan}\sqrt{y}x{|}_{0}^{1}=\frac{\mathrm{arctan}\sqrt{y}}{\sqrt{y}}.\end{array}$

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