Proving the Alternate Series Test I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott. I am finding it difficult to come up with a proof for the Alternating Series Test for an infinite series. I only have some initial ideas, so I'll put those down.
I'd like to ask if somebody could help, by providing any hints (please do not write the complete proof), on how to think about this.
Exercise 2.7.1

Proving the Alternating Series Test amounts to showing that the sequence of partial sums
\begin{align*}
 s_n = a_1 - a_2 + a_3 - \ldots \pm a_n
\end{align*}
converges. Different characterizations of completeness lead to different proofs.


(a) Produce the Alternating Series Test by showing that $(s_n)$ is a Cauchy sequence.


(b) Supply another proof for this result using the Nested Interval Property.


(c) Consider the sequences $(s_{2n})$ and $(s_{2n+1})$, and show how the Monotone Convergence Theorem leads to a third proof for the Alternating Series Test.

Proof.
(a)Let $(a_n)$ be a sequence satisfying
(i) $a_1 \ge a_2 \ge a_3 \ge \ldots \ge a_n \ge a_{n+1} \ge$
(ii) $(a_n) \to 0$
Our claim is that, the alternating series $\sum_{n=1}^{\infty}(-1)^n a_n$ converges.
Let $(s_n)$ be sequence of the partial sums, where
\begin{align*}
s_n = a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + \ldots + (-1)^{n+1}a_n
\end{align*}
We are interested to show that $(s_n)$ is a Cauchy sequence. Let's see if can find a simple expression for $\lvert s_n - s_m \vert$. If $n > m$,
\begin{align*}
\lvert (s_n - s_m) \rvert &= \lvert (-1)^{m+2}a_{m+1} + (-1)^{m+3}a_{m+2} + (-1)^{n+1}a_{n} \rvert\\
&= \lvert {a_{m+1} - a_{m+2} + a_{m+3} - \ldots + (-1)^{n-m+1}a_n} \rvert
\end{align*}
A consequence of condition(i) is that $(a_n)$ is a decreasing sequence and has a lower bound zero, so $a_n \ge 0$ for all $n \in \mathbf{N}$. Since $a_{m+2} \ge 0$, $a_{m+4} \ge 0$, $a_{m+6} \ge 0, \ldots$ we can write,
\begin{align*}
\lvert (s_n - s_m) \rvert &= \lvert {a_{m+1} - a_{m+2} + a_{m+3} - \ldots + (-1)^{n-m+1}a_n} \rvert\\
&\le \lvert {a_{m+1} + a_{m+3} + a_{m+5} + \ldots + a_{n}} \rvert \\
&\le \vert {a_{m+1} + a_{m+1} + a_{m+1} + \ldots + a_{m+1}} \rvert
\end{align*}
But this gives me a variable number of terms (whereas I need to prove that the distance $\lvert s_n - s_m \rvert$ becomes smaller than fixed $\epsilon$ or $k\epsilon$. I am not quite sure, where to go from here. I know that the original sequence $(a_n)$ converges, so given any $\epsilon > 0$, $\exists N \in \mathbf{N}$, such that $\lvert a_n \rvert < \epsilon$ for all $n \ge N$.
(b) I am thinking of constructing a sequence of closed nested intervals $I_1 \supseteq I_2 \supseteq I_3 \supseteq \ldots \supseteq I_n \supseteq I_{n+1} \supseteq \ldots$, in such a way that, $I_1=\{s_1,s_2,s_3,\ldots\}$, $I_2 = I_1 - \{s_1\}$ and in general $I_{k+1} = I_k - \{s_k\}$, as I am interested in the tail of the sequence of partial sums. Then, I can apply the nested interval property to show that $\bigcup_{k\ge 1}I_k$ is not empty.
The one thing I don't know, is whether $(s_n)$ is bounded. If and only if this set is bounded, one may construct a closed interval. I think this is important for the construction.
(c) Again, I need some bounds for $(s_n)$, I guess.
 A: Consider  $s_{2n} = a_1 - (a_2 - a_3) - \cdots -(a_{2n-2} - a_{2n-1} ) - a_{2n} \leq a_1 $ and it is easy to observe that $s_{2n}$ is increasing . So it is bounded and monotone hence converges and let the limit be $s$. Further we have $s_{2n+1} = s_{2n} + a_{2n+1}$. Since $a_n \to 0$ as $n \to \infty$ we have $\lim s_{2n+1} = \lim s_{2n} = s$.
A: $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
For completeness, I add the proof using NIP(nested interval property).
Given that $a_1 \ge a_2 \ge a_3 \ge \ldots $
\begin{align*}
 s_n &= a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + \ldots + (-1)^{n+1}a_n\\
  &= a_1 - (a_2 - a_3) - (a_4 - a_5) - (a_6 - a_7) - \ldots \\
  &\le a_1 = s_1
\end{align*}
And, moreover,
\begin{align*}
 s_n &= (a_1 - a_2) + (a_3 - a_4) + (a_5 - a_6) + \ldots + (-1)^{n+1}a_n\\
  &\ge 0
\end{align*}
So, $0 \le s_n \le s_1$ for all $n \in \mathbf{N}$.
Also, consider the subsequences,
\begin{align*}
 (s_{2n}) &= (s_2,s_4,s_6,s_8,\ldots,s_{2n-2},s_{2n})   \\
 (s_{2n-1}) &= (s_1,s_3,s_5,s_7,\ldots,s_{2n-3},s_{2n-1})
\end{align*}
We observe:
\begin{align*}
 s_{2} &= a_1 - a_2\\
 s_{4} &= (a_1 - a_2) + (a_3 - a_4) \\
 s_{6} &= (a_1 - a_2) + (a_3 - a_4) + (a_5 - a_6)
\end{align*}
Thus, $(s_{2n})$ is a monotonically increasing sequence.
\begin{align*}
 s_{1} &= a_1\\
 s_{3} &= a_1 - (a_2 - a_3)\\
 s_{6} &= a_1 - (a_2 - a_3) - (a_4 - a_5)
\end{align*}
Thus, $(s_{2n-1})$ is a monotonically decreasing sequence.
Let $S := [0,s_1]$. Next, let $I_1$ be the closed interval that contains $s_2$ and $s_3$. In general,
\begin{align*}
 I_{k} = [s_{2k},s_{2k+1}]
\end{align*}
By the nested interval property, $\bigcap_{k\ge1} I_k \ne \emptyset$ and contains atleast one element $x$.

Define $A :=\{s_2,s_4,s_6,\ldots\}$ and $B := \{s_1,s_3,s_5,s_7,\ldots\}$. Every $s_{2n+1} \in B$ is an upper bound for $A$ and every $s_{2n} \in A$ is a lower bound for $B$. Since, these are bounded subsets of $\mathbf{R}$, we are justified in setting $x = \sup A = \inf B$. Thus, we have a candidate for the limit of the alternating series. Our claim then is, $\lim s_{2n} = x = \lim s_{2n+1}$.
Since $x = \sup A$, for any $\epsilon > 0$, $x - \epsilon$ is not an upper bound for A. Thus, there exists a point $s_{2N} \in A$, such that $x - \epsilon < s_{2N}$. Therefore, $\absval{s_{2N} - x} < \epsilon$. A similar argument can be made for $B$.
