Prove the uniform convergence of ${\frac{1}{(1+ {\frac{y^2}{n}})^n}}$

Question

How can I prove the uniform convergence for these two tasks:

1. $${\frac{1}{(1+ {\frac{y^2}{n}})^n}} ⇉ e^{-y^2} $$ I understand that the function on the right is the limit function, because: $$\lim _{{x\to \infty }}\left(1+{\frac {1}{x}}\right)^{x}=e.$$ but how can I prove exactly uniform convergence?
2. $$∫_0^∞ \frac{dy}{(1+ {\frac{y^2}{n}})^n} ⇉$$

Answer 1

Let $f_n(x)$ be the sequence

$$f_n(y)=\left(1+\frac{y^2}{n}\right)^{-n}-e^{-y^2}\tag1$$

Since $\left(1+\frac{y^2}{n}\right)^{-n}$ monotonically decreases to $e^{-y^2}$ (See this answer as a reference), we see from $(1)$ that $f_n(y)\ge0$ for all $y\ne 0$.

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Setting $f_n'(y)=0$, we find the maximum value of $f_n(y)$ occurs at some value $y^*>0$ (and $-y^*$) for which

$$e^{-{y^*}^2}=\left(1+\frac{{y^*}^2}{n}\right)^{-n-1}\tag2$$

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Using $(2)$, reveals that the maximum value of $f_n(y)$ is given by

$$\begin{align} \max_y(f_n(y))&=f_n(\pm y^*)\\\\ &=\left(1+\frac{{y^*}^2}{n}\right)^{-n}-e^{-{y^*}^2}\\\\ &=e^{-{y^*}^2}\left(1+\frac{{y^*}^2}{n}-1\right)\\\\ &=\frac{{y^*}^2e^{-{y^*}^2}}n\\\\ &\le \frac1{ne} \end{align}$$

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Hence, we find that

$$\sup_{y}\left|\left(1+\frac{y^2}{n}\right)^{-n}-e^{-y^2}\right|\le \frac{1}{ne}\to 0$$

and conclude that the sequence $\left(1+\frac{y^2}n\right)^n$ converges uniformly to $e^{-y^2}$ for $y\in \mathbb{R}$ as was to be shown!

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Note that $f_n(y)$ is continuous in $y$ for all $n$ and that $\int_0^\infty f_n(y)\,dy$ converges uniformly for $n\ge 1$. Then, the Moore-Osgood Theorem guarantees that we can interchange limits so that

$$\lim_{n\to\infty}\lim_{L\to\infty}\int_0^L f_n(y)\,dy=\lim_{L\to\infty}\lim_{n\to\infty}\int_0^L f_n(y)\,dy$$

Inasmuch as $f_n(y)$ converges uniformly for all $y$, we have

$$\begin{align} \lim_{L\to\infty}\lim_{n\to\infty}\int_0^L f_n(y)\,dy&=\lim_{L\to\infty}\int_0^L \lim_{n\to\infty} f_n(y)\,dy\\\\ &=\lim_{L\to\infty}\int_0^L e^{-y^2}\,dy\\\\ &=\frac{\sqrt\pi}{2} \end{align}$$