Uniform convergence of $f_n(x) = \left(1 + \frac{x}{n}\right)^n$ when calculating limit Calculate$$
\lim_{n \rightarrow \infty} \int_0^1 \left(1 + \frac{x}{n}\right)^ndx
$$
My attempt - if
$$
f_n(x) = \left(1 + \frac{x}{n}\right)^n
$$
converged uniformly for all $x \in [0,1]$ then I could swap integral with limes and solve it:
$$
\lim_{n \rightarrow \infty} \int_0^1 \left(1 + \frac{x}{n}\right)^ndx = 
 \int_0^1 \lim_{n \rightarrow \infty}\left(1 + \frac{x}{n}\right)^ndx =
 \int_0^1 e^x dx = e^x|_{0}^{1} = e - 1
$$
I must then prove that $f_n(x)$ is indeed uniformly convergent. I already know that
$f_n(x) \rightarrow e^x$. If $f_n(x)$ converges uniformly then for each epsilon the following statement must hold
$$
\sup_{x \in [0,1]} \left|f_n(x) - f(x)\right| < \epsilon
$$
How can I prove this?
 A: Alternative approach (without uniform convergence): let $t= \frac{x}{n}$ then
$$\begin{align*}\int_0^1 \left(1 + \frac{x}{n}\right)^ndx&=
n\int_0^{1/n} (1+t)^ndt\\
&=n\left[\frac{(1+t)^{n+1}}{n+1}\right]_0^{1/n}
\\&=
\frac{n}{n+1}\left(\left(1+\frac{1}{n}\right)^{n+1}-1\right)\\&\to e-1.
\end{align*}$$
A: You can use Dini's theorem.
On a compact set $K$, if a sequence of continuous functions $\langle f_n(x) \rangle$
$a)$ is monotone in $n$ for each $x \in K$
$b)$ converges pointwise to a continuous function of $x \in K$
then the convergence is uniform.
A: To show uniform convergence, you can just estimate the error term:
Note that
$$
n\log(1+x/n)=x-\frac1{2n}x^2+\frac1{3n^2}x^3-\dots
$$
is an alternating series with each term smaller in magnitude (since $x\in[0,1]$), so
$$
\lvert x-n\log(1+x/n)\rvert\leq\frac1{2n}x^2
$$
So for all $x\in[0,1]$, we have
\begin{align*}
0&\leq e^x-\left(1+\frac{x}{n}\right)^n\\
&=e^x-e^{n\log(1+x/n)}\\
&\leq\lvert x-n\log(1+x/n)\rvert\cdot\sup\{e^\xi\mid n\log(1+x/n)\leq\xi\leq x\} &&\text{(MVT)}\\
&\leq \frac{x^2}{2n}e^x\leq\frac{e}{2n}
\end{align*}
So the convergence is uniform.
A: If $t_n(x)=\left(1+\frac{x}{n}\right)^n$, then, expanding $t_n,$ 
$t_n(x)=1+x+\frac{x^2}{2!}\left(1-\frac{1}{n}\right)+\cdots+\frac{x^{n}}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\times \cdots \times \left(1-\frac{n-1}{n}\right)$
noting that $0\le x\le 1,$ and comparing this with $s_n=\sum^n_{k=0}\frac{1}{k!}$, we have $t_n(x)\le s_n\le \limsup s_n=e$ and so the result follows by the dominated convergence theorem.
A: The uniform convergence follows from two exercises:


*

*For $x\in [0,1],$ $n\ln(1+x/n) - x\to 0$ uniformly on $[0,1].$

*If $f_n\to f$ unifomly on a set $E,$ and $f$ is bounded, then $e^{f_n}\to e^f$ uniformly on $E.$
To prove 1., rewrite the expression as
$$x\left (\frac{\ln(1+x/n)-\ln(1)}{x/n}-1\right )$$
and use the fact that $\ln'(1)=1.$ For the proof of 2., note $|e^{f_n}- e^f| = e^f|e^{f_n-f}-1|.$
A: On $[0, 1]$,
$$
\left(1 + \frac x n \right)^n \le
e^x \le
e^1 =
e
$$
So that by the Dominated Convergence Theorem,
$$
\lim_{n \rightarrow \infty} \int_0^1 \left(1 + \frac x n \right)^n\ dx=
\int_0^1 \lim_{n \rightarrow \infty} \left(1 + \frac x n \right)^n\ dx =
\int_0^1 e^x\ dx =
e - 1
$$
