I need to prove $f_n(x)=(1-x/n)^n$ converges on non-negative reals. I can easily prove it converges to $f(x)=e^{-x}$, but it is unclear to me whether this convergence is pointwise or uniform.
I have attempted the proof by taking the Taylor expansion of $f$ and do the binomial expansion of $f_n$, and then subtract.
\begin{equation} |f(x)-f_n(x)|=|\sum_{k=0}^{\infty}(-1)^k\frac{x^k}{k!}-\sum_{k=0}^{n}(-1)^k\binom{n}{k}(\frac{x}{n})^k|\\ =|\sum_{k=0}^{n}(-1)^k\frac{x^k}{k!}-(-1)^k\binom{n}{k}(\frac{x}{n})^k+O(\frac{1}{n^{k+1}})| \end{equation}
Toss out the big O term for now, since it is just part of the Taylor expansion of $e^{-x}$, with $k>n$. I tried manipulate the terms.
\begin{equation} =|\sum_{k=0}^{n}(-1)^k\frac{x^k}{k!}(1-\frac{n(n-1)...(n-k+1)}{n^k})|\\ =|\sum_{k=0}^{n}(-1)^k\frac{x^k}{k!}\frac{n^k-n(n-1)...(n-k+1)}{n^k}|\\ \end{equation}
For any non-negative real $x$, each term in the summation clearly converges as we take $n$ large enough. But given a fixed $\epsilon>0$, I can't seem to find $N\in \mathbb{Z^+}$ such that such that $\forall n>N$, $|f(x)-f_n(x)|<\epsilon$. In other word, I have trouble proving the speed of the convergence does not depend on $x$.
Any help would be greatly appreciated, as I really would like to understand this. Thank you in advance!
The convergence is not uniform: if it was, then for all $\varepsilon>0$ there exists an $n_0\in\mathbb N$ such that, if $n\geq n_0$, then $|f_n(x)-e^{-x}|<\varepsilon$ for all $x\geq 0$, so $|f_n(2n)-e^{-2n}|<\varepsilon$ for all $n\geq n_0$. Therefore, the sequence $(f_n(2n)-e^{-2n})$ converges to $0$. But, $$f_n(2n)-e^{-2n}=\left(1-\frac{2n}{n}\right)^n-e^{-2n}=(-1)^n-e^{-2n},$$ which does not converge to $0$, and this is a contradiction.