I'm trying to prove that following:
Let $ \{ b_n \}_{n=1}^{\infty}$ be a positive, decreasing sequence and let $s_n = \sum_{k=1}^{\infty} (-1)^{k-1} b_k$. Prove that $\{s_{2n} \}$ is monotone and bounded.
I understand the problem intuitively, but my proof-writing has historically been a little weak. My attempt:
Since $b_n$ is decreasing, it follows that for arbitrary $n \in \mathbb{N}$,
$$b_{2n+1} > b_{2n+2} \iff 0 < b_{2n+1} - b_{2n+2}$$
Adding the quantity $\sum_{k=1}^{2n} (-1)^{k-1} b_k$ to both sides, we obtain \begin{align*} \sum_{k=1}^{2n} (-1)^{k-1} b_k &< \bigg[ \sum_{k=1}^{2n} (-1)^{k-1} b_k \bigg] + b_{2n+1} - b_{2n+2} \\ &= \sum_{k=1}^{2n+2} (-1)^{k-1} b_k \\ \end{align*} Or equivalently,
$$s_{2n} < s_{2n+2} $$
Thus, $s_{2n}$ is monotonically increasing. Now, we claim that $s_{2n}$ is bounded above by $b_1$. Again, since $b_n$ is decreasing, we conclude that for any $n$, \begin{align*} s_{2n} &= b_1 - b_2 + b_3 - \dots -b_{2n-2} + b_{2n-1} - b_{2n} \\ &< b_1 - b_3 + b_3 - \dots -b_{2n-1} + b_{2n-1} - b_{2n} \\ &= b_1 - b_{2n} \\ &< b_1 \end{align*} Note that the final step follows from the fact that $b_1 = max\{b_n\}$. Further, \begin{align*} s_{2n} &= b_1 - b_2 + b_3 - b_4 \dots + b_{2n-1} - b_{2n} \\ &> 0 + 0 + \dots + 0 \\ &= 0 \\ \end{align*} so $s_{2n}$ is bounded below by $0$. Thus, $s_{2n}$ is monotone and bounded.
Is this proof thorough enough?