Uniform Convergence of an Exponential Sequence of Functions How can I show that for every $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that
$$\left|f_n(x)-f(x)\right|=\left|\left(\frac{x}{n}+1\right)^n-e^x\right|<\epsilon$$
whenever $n\geq N$ and $x\in\left[-A,A\right]$? By the way, $n\in\mathbb{N}$.
In a previous exercise, I was able to show that $f_n$ does indeed converge pointwise to $f$. However, I have been stuck for hours trying to prove uniform convergence. Would anyone lend me a hand? Thanks in advance.
 A: Write
$$(\frac{x}{n}+1)^n=\exp\left( n \log(1 + \frac{x}{n})\right).$$
Assuming that $N>2A$ and hence $|x/n|<\frac12$, $\log (1+\frac{x}{n})$ can be expanded in a Taylor series with remainder, giving
$$\log(1 + \frac{x}{n})=\frac{x}{n}-\frac{1}{(1+(y/n))^2} \frac{x^2}{2n^2},\qquad |y|\le |x|$$
$$=\frac{x}{n}-\theta \frac{x^2}{2n^2},\qquad \theta\in [0,4].$$
Substituting this into the first equation gives$$(\frac{x}{n}+1)^n = \exp\left(x-\theta \frac{x^2}{2n}\right)$$
so
$$\left|(\frac{x}{n}+1)^n-e^x\right|=e^x (1-\exp -\theta \frac{x^2}{2n}).$$
The right-hand side of this is no bigger than $e^A (1-e^{-4A^2/2n})$, so uniform convergence follows.
A: You (might) know that
$$e^x=1+x+\frac{x^2}{2!}+\cdots$$
and that
$$\left(1+\frac{x}{n} \right)^n=1+x+\frac{n(n-1)}{n^2}\frac{x^2}{2!}+\frac{n(n-1)(n-2)}{n^3}\frac{x^3}{3!}+\cdots$$
So you get
$$\left|\left(1+\frac{x}{n} \right)^n-e^x \right|\leqslant \left| \frac{n(n-1)}{n^2}-1 \right|\frac{x^2}{2!}+ \left| \frac{n(n-1)(n-2)}{n^3}-1 \right|\frac{x^3}{3!}+\cdots$$
All the expressions in  $n$ will be rational and of the form $\dfrac{P(n)}{Q(n)}$, where the degree of $P$ will be one less than $Q$. For example, the first two are:
$$ \left| \frac{1}{n} \right|$$
$$ \left| \frac{3n^2-2n}{n^3} \right|$$
They will all behave like $\dfrac{C}{n}$ for some constant $C$ for large $n$. It shouln't be too hard for you to show that this difference can be made arbitrarely small for sufficiently large $n$.
