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 $kepsilon$. 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 $bigcup_{kge 1}I_k$ 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.