Why alternating power series can be truncated to created upper and lower bounds for functions? I often see functions that can be represented by alternating power series like this
\begin{align}
f(x) =\sum_{i=0}^\infty(-1)^i a_i x^i,
\end{align}
being upper and lower bounded by a truncated power series. For example, the function above can be bounded as follows:
\begin{align}
a_0 - a_1 x\le f(x)\le a_0 - a_1 x + a_2 x^2
\end{align}
However, I am curious about why is this so? I think it has something to do with how fast the approximate error decreases.
 A: There is a theorem about constant alternating series $\sum_{k=0}^\infty (-1)^k c_k$ with positive $c_k$ monotonically decreasing to $0$. Such series are automatically convergent to a finite sum $s\in{\mathbb R}$. Furthermore the even partial sums $s_{2m}:=\sum_{k=0}^{2m}(-1)^k c_k$ are all larger than $s$, and the odd partial sums $s_{2m+1}$ are all smaller than $s$. This can be expressed  in the following way: The truncation error $s-s_n$ is smaller in absolute value than the first neglected term, and has the same sign as this term.
If you have a power series $f(x):=\sum_{k=0}^\infty (-1)^k a_kx^k$ with positive $a_k$ this principle can be applied for positive $x$-values   small enough to guarantee that $a_{k+1}x^{k+1}<a_k x^k$, or $a_{k+1}x<a_k$, for all $k$. 
Consider as an example the series
$$e^{-x}=\sum_{k=0}^\infty(-1)^k{x^k\over k!}\ .$$
If, e.g., $x=5$ then the stated condition is only fulfilled for $k\geq6$. It follows that the principle in question is only applicable for this $x$  if you truncate the series after the sixth term.
