Uniform convergence,quick proof I am trying to prove the uniformly convergence to exp(-x):
$$f_n(t) = \left\{ \begin{array}{ll}
        \left(1-\frac{x}{n}\right)^n & \mbox{if } x\in [0,n] \\
        0 & \mbox{if } x\geq n
    \end{array}
\right.
$$
But my proof is long and ugly so I imagine there is a better demonstration. If anyone knows a quick proof...
Thanks in advance
 A: Since $1-x \le e^{-x}$, we have that $(1-\frac xn)^{n} \le (e^{x/n})^n = e^{-x}$.
Given $\epsilon>0$, if $x > \ln(1/\epsilon)$, then $e^{-x} < \epsilon$.  So we need only prove uniform convergence on finite intervals.
So fix $L>0$.  Find $N \ge L$ such that for $y \ge N$, $(1-\frac Ly)^y \ge e^{-L}(1 - \epsilon)$.  Then for $x \in [0,L]$, and $n \ge N$, since $Ln/x \ge N$, we have $(1-\frac xn)^n = ((1-\frac{L}{Ln/x})^{Ln/x})^{x/L} \ge (e^{-L}(1 - \epsilon))^{x/L} \ge e^{-x}(1-\epsilon)$.
Note we use that $(1-x)^\alpha \ge 1-x$ if $0\le x \le 1$ and $0 \le \alpha \le 1$ in the last step.
A: First proof. As noted by Stephen Montgomery-Smith, it is enough to prove the uniform convergence on segments $[0,K]$ for all $K > 0$.
Let us recall that if two sequences $(a_i)$ and $(b_i)$ satisty $|a_i| \leq 1$ and $|b_i|\leq 1$ for all $i$ then $\left|\prod_{i=1}^n a_i - \prod_{i=1}^n b_i\right| \leq \sum_{i=1}^n |a_i-b_i|$. Using this with $a_i = (1-x/n)$ and $b_i = e^{-x/n}$ and the first-order Taylor expansion of $e^{-x}$, we obtain that for all $x \in [0,K]$,
$$
\left|f_n(x) - e^{-x}\right| \leq n \left|e^{-x/n} -1 + \frac{x}{n}\right| \leq n\frac{x^2}{2n^2} \leq \frac{K^2}{2n}
$$
and the uniform converge is now obvious.
Second proof.
Using the inequalities $e^{-u/(1-u)} \leq 1-u \leq e^{-u}$ for $0 \leq u \leq 1$ with $u = \frac{x}{n}$, we obtain again for all $x \in [0,K]$,
$$
0 \leq e^{-x} - f_n(x) \leq e^{-x}\left(1-e^{-\frac{x^2}{n-x}}\right) \leq e^{-0}\frac{K^2}{n-K} \xrightarrow[n\to\infty]{} 0.
$$
Third proof. Apply the following lemma due to George Pólya : given a sequence of functions $f_n \colon [0,\infty] \to \Bbb R$ and $f\colon [0,\infty] \to \Bbb R$ such that


*

*the function $f$ is continuous,

*for all $n$, the function $f_n$ is decreasing,

*for all $x \in [0,\infty]$, the sequence $f_n(x))$ converges to $f(x)$ as $n \to \infty$;


then $f_n$ converges uniformly to $f$ on $[0,\infty]$.
