From pointwise to uniform convergence on compact sets

Question

It is a well-known fact that the sequence of functions $\{(1+\frac{x}{n})^n\}_{n\in\mathbb{N}}$ converges pointwise to $e^x$ on $\mathbb{R}$. How to prove that this defines a uniform convergence on compact sets?

My attempt is analogous to the proof of Dini's theorem:

Let $r\in\mathbb{R}$ and let $\varepsilon\in\mathbb{R}$ be positive. For all $n\in\mathbb{N}$ we define $$ I_n=\left\{x\in [-r,r]\mid \left|e^x-\left(1+\frac{x}{n}\right)^n\right|<\varepsilon \right\}, $$ then, by continuity of the functions involved, $\{I_n\}_{n\in\mathbb{N}}$ is a sequence of open sets. Such sequence is ascending (why?) and, by the pointwise convergence stated above, it covers $[-r,r]$.

By compactness, there exists $N\in\mathbb{N}$ such that $\bigcup_{n=0}^N I_n=[-r,r]$ and the conclusion easily follows.

Anyway, this method seems quite artificious. Also, hot to prove that $\{I_n\}_{n\in\mathbb{N}}$ is ascending? Unfortunately, $e^x-\left(1+\frac{x}{n}\right)^n$ leads to difficult computations. Is there a more direct proof, for instance, to find that $N$?

Answer 1

Hint: It may be simpler to work with the logarithm. To that end, note that if $f_n \to f$ uniformly on $[a,b],$ and $f$ is bounded, then $e^{f_n} \to e^f$ uniformly on $[a,b].$ So if we first prove that $n\ln (1+x/n) \to x$ uniformly on $[-r,r],$ then we can exponentiate back and apply the exercise.

Answer 2

We can omit finitely many $n$ from the sequence $(I_n)_n.$ If $C$ is a compact subset of $\Bbb R$ let $n_0\in \Bbb N$ such that $\forall x\in C\;(|x|\leq n_0-1).$

For $x\in C$ and $y\geq n_0$ the function $$F_x(y)=y\log (1+x/y)=\log((1+x/y)^y)$$ is increasing in $y,$ which can be seen by examining $dF_x(y)/dy$ and $d^2F_x(y)/dy^2.$ The computations are easy.

So for $x\in C$ the sequence $(\;(1+x/n)^n\;)_{n\geq n_0}$ increases monotonically to $e^x.$ Therefore $I_n(C,\epsilon)\subset I_{n+1}(C,\epsilon)$ for $n\geq n_0.$