Comparing $e^{-x}$ to its partial sums. Context: Many definite integral problems induce inequalities by involving basic inequalities (such as $\sin x, \cos x$ are each always greater than all their partial sums). So, what about $e^{-x}$?
First off, I know that $$e^x>\sum_{i=0}^k\frac{x^i}{i!}$$ for any natural number $k>0$ and $x>0$.
But what about $e^{-x}?$ Firstly, I feel that the comparison should alternate: $e^{-x} <S_k=\displaystyle\sum_{i=0}^k\frac{(-x)^i}{i!}$ when k is even (as the next largest term is being subtracted from the partial sum to get $e^{-x}$) and $e^{-x} >S_k$ for odd k.
But then again I feel that $S_k$ is a partial sum, so it should intuitively be less than $e^{-x}$.
How does $e^{-x}$ compare to its partial sums?
Please keep it to high-school level.
 A: Nice question! Your guess is correct. Here's a fun proof: to show that a function $f(x)$ satisfies $f(x) \ge 0$ for $x \ge 0$ it's enough to show that $f(0) \ge 0$ and $f'(x) \ge 0$ for $x \ge 0$ (e.g. by the mean value theorem). Consider the sequence of functions
$$f_k(x) = (-1)^{k+1} \left( e^{-x} - \sum_{i=0}^k \frac{(-x)^i}{i!} \right), k \ge 0$$
We want to show that $f_k(x) \ge 0$ for $x \ge 0$. We can do this by induction on $k$: we have $f(0) = 0$ for all $k$, and (this is the key computation!) $f_k'(x) = f_{k-1}(x)$, so once we know that $f_0(x) = 1 - e^{-x} \ge 0$ for $x \ge 0$ we're done by induction. The same induction gives the strict inequality $f_k(x) > 0$ for $x > 0$ for all $k$.
A: HINT:
Show that the Taylor series of $e^x\cdot  S_k(x)$ is of the form
$$e^x \cdot S_k(x) = 1 + (-1)^k \cdot x^{k+1}\cdot  (\textrm{ series with positive coefficients})$$
$\bf{Added:}$ Interestingly enough, a similar thing happens with the other basic inequalities. For instance, the Taylor series of
$$\frac{1 - \frac{x^2}{2} + \frac{x^4}{24}}{\cos x}$$
has positive coefficients.
