This is used in an elementary proof that $e$ is irrational. I can prove this, but what I am doing is not particular nice looking. In the proof, the author says this is obvious but I can't seem to write it out so simply.
\begin{equation} S = \sum_{k = 0}^\infty (-1)^k a_k \end{equation}
where $a_k > a_{k + 1} > 0$ and $a_k \rightarrow 0$. So, I want to show that
\begin{equation} a_0 - a_1 < S < a_0 \end{equation}
For a convergent series with decreasing terms (which is almost certainly what the author's talking about, because the claim isn't true in general!) observe that every odd partial sum like $a_0 - a_1 + a_2 - a_3 + a_4$ can be written as $a_0$ minus a collection of nonnegative numbers, like $a_0 - (a_1 - a_2) - (a_3 - a_4)$, hence is no greater than $a_0$. Do the same thing for even partial sums compared to $a_0 - a_1$, and you're done.
The even partial sums are positive and decreasing, so they converge. The odd partial sums are bounded from above by $a_0$ and increasing, so they also converge. The even and odd partial sums must converge to the same limit because their difference $S_{2n} - S_{2n+1} = a_{2n+1}$ converges to zero. Hence the entire series converges, and now John Hughes' argument proves the bounds on the limit.
