Uniform convergence of $f_n(x) = \left(1 + \frac{x}{n}\right)^n$ when calculating limit

Question

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?

Answer 1

Alternative approach (without uniform convergence): let $t= \frac{x}{n}$ then $$\begin{align*}\int_0^1 \left(1 + \frac{x}{n}\right)^ndx&= n\int_0^{1/n} (1+t)^ndt\\ &=n\left[\frac{(1+t)^{n+1}}{n+1}\right]_0^{1/n} \\&= \frac{n}{n+1}\left(\left(1+\frac{1}{n}\right)^{n+1}-1\right)\\&\to e-1. \end{align*}$$

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

The uniform convergence follows from two exercises:

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

2. 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 $|e^{f_n}- e^f| = e^f|e^{f_n-f}-1|.$

Answer 6

On $[0, 1]$, $$ \left(1 + \frac x n \right)^n \le e^x \le e^1 = e $$

So that by the Dominated Convergence Theorem,

$$ \lim_{n \rightarrow \infty} \int_0^1 \left(1 + \frac x n \right)^n\ dx= \int_0^1 \lim_{n \rightarrow \infty} \left(1 + \frac x n \right)^n\ dx = \int_0^1 e^x\ dx = e - 1 $$