upper bound of exponential function I am looking for a tight upper bound of exponential function (or sum of exponential functions):
$e^x<f(x)\;$ when $ \;x<0$ 
or
$\displaystyle\sum_{i=1}^n e^{x_i} < g(x_1,...,x_n)\;$ when $\;x_i<0$ 
Thanks a lot!
 A: Since you suggest in the comments you would like a polynomial bound, you can use any even Taylor polynomial for $e$.
Proposition. $\boldsymbol{1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}}$ is an upper bound for $\boldsymbol{e^x}$ when $\boldsymbol{n}$ is even and $\boldsymbol{x \le 0}$.
Proof.
We wish to show $f(x) \ge 0$ for all $x$, where $f: (-\infty, 0] \to \mathbb{R}$ is the function defined by
$f(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} - e^x.$
Since $f(x) \to \infty$ as $x \to - \infty$,
$f$ must attain an absolute minimum somewhere on the interval $(-\infty, 0]$.


*

*If $f$ has an absolute minimum at $0$, then for all $x$,
$f(x) \ge f(0) = 1 - e^0 = 0$, so we are done.

*If $f$ has an absolute minimum at $y$ for some $y < 0$, then $f'(y) = 0$.
But differentiating,
$$
f'(y) = 1 + y + \frac{y^2}{2!} + \cdots + \frac{y^{n-1}}{(n-1)!} - e^y
= f(y) - \frac{y^n}{n!}.
$$
Therefore, for any $x$,
$$
f(x) \ge f(y) = \frac{y^n}{n!} + f'(y) = \frac{y^n}{n!} > 0,
$$
since $n$ is even.
$\square$

Keep in mind that any polynomial upper bound will only be tight up to a certain point, because the polynomial will blow up to infinity as $x \to -\infty$.
Also note, the same proof shows that the Taylor polynomial is a lower bound for $e^x$ when $n$ is odd and $x \le 0$.
A: For $x < 0$ and $n \in \mathbb{N}$,
$$
0 < \sum_{k=0}^{n} \frac{(-x)^k}{k!} < e^{-x}
$$
so
$$
e^x < 1\left/\sum_{k=0}^{n} \frac{(-x)^k}{k!}\right..
$$
A: How about $e^{-x}\leq c(\gamma)(\frac{1}{1+x})^\gamma$ for non-negative $x$?
