$(1-\frac{x}{n})^n\lt \exp(-x)$ Why is the following inequality true: if $x \geq 0$ then $(1-\frac{x}{n})^{n} \leq e^{-x}$ ? here $n$ is a positive integer. Is there a quick way to see this?
 A: You might want to prove this for $0 \leq x \leq n$. Else it might not be always true. (For instance, consider the case when $x$ is really large and $n$ is even)
Consider the function $f(y) = e^{y} \times (1-y)$. Note that $f(0) = 1$. 
We will now prove that $f(y)$ is a decreasing function $\forall y \geq 0$.
$\frac{df(y)}{dy} = e^{y} - ye^{y} - e^{y} = -ye^{y} \leq 0$, $\forall y \geq 0$.
Hence $f(y)$ is a decreasing function $\forall y \geq 0$.
So, we have $f(y) \leq f(0) = 1$, $\forall y \geq 0$.
Hence, we get $e^{y} \times (1-y) \leq 1$, which implies $(1-y) \leq e^{-y}$.
Replace $y = \frac{x}{n}$ and raise both sides to the $n^{th}$ power where $n \in \mathbb{R}^{+}$. The inequality remains the same since $0 \leq x \leq n$
So, we get $(1-\frac{x}{n})^n \leq e^{-x}$.
It is also useful to see that $\displaystyle \lim_{n \rightarrow \infty} (1-\frac{x}{n})^n = e^{-x}$ from below.
A: consider the case where $x = 4$ and $n=2$. Here, LHS > RHS. So the inequality does not always hold.
A: You can look at it this way:
$(1 - \frac{x}{n})^n = \sum_{k=0}^{n} C(n,k) (-1)^k \frac{x^k}{n^k}$
and
$e^{-x} = \sum_{k=0}^{+\infty} (-1)^k \frac{x^k}{k!}$
Is it clearer now ?
(hint: one of the sums is finite, the other not, and one of the terms grows quicker than the other)
A: First thing: you need $n$ to be even for this if  1 - x/n is negative.
Your statement is the same as $1 - {x \over n} \leq e^{-{x \over n}}$. Letting $t = {x \over n}$ this follows from $1 - t \leq e^{-t}$. This can be proven using calculus.
