Let \(a_0, a_1, a_2, \ldots\) be a decreasing sequence of positive real numbers. Prove that for every nonnegative integer \(m\), \[0 \le \sum_{i=0}^{m}(-1)^i a_i \le a_0.\] (Hint: \(\sum_{i=0}^{m+1}(-1)^i a_i = a_0 - \sum_{i=0}^{m}(-1)^i a_{i+1}\).)
Solution
Let \[ S_m=\sum_{i=0}^m(-1)^ia_i. \] Because \((a_i)\) is decreasing and positive, each difference \(a_{2j}-a_{2j+1}\) is nonnegative.
If \(m=2r+1\) is odd, then \[ S_m=(a_0-a_1)+(a_2-a_3)+\cdots +(a_{2r}-a_{2r+1})\ge0. \] If \(m=2r\) is even, then \[ S_m=(a_0-a_1)+\cdots +(a_{2r-2}-a_{2r-1})+a_{2r}\ge0. \] Thus \(S_m\ge0\) in all cases.
For the upper bound, if \(m=2r\), then \[ S_m=a_0-(a_1-a_2)-\cdots-(a_{2r-1}-a_{2r})\le a_0. \] If \(m=2r+1\), then \[ S_m=a_0-(a_1-a_2)-\cdots-(a_{2r-1}-a_{2r})-a_{2r+1}\le a_0. \] Therefore \[ 0\le S_m\le a_0 \] for every \(m\ge0\).