# Proving the Alternate Series Test with Monotone Convergence

Our book is asking us to prove the Alternate Series Test, but use the method it describes (Making this question different from others already posted).

$s_n = a_1 - a_2 + a_3 - a_4 + a_5 - ...$

The book asks us to look at the sub-sequences $s_{2n}$ and $s_{2n+1}$ and specifically asks us to use the monotone convergence theorem to prove this theorem.

I need some confirmation that this solution works (or what I should change).

$\textbf{My attempt}$

Let $({a_n})$ be a sequence that is monotone decreasing and converging to $0$.

Let $s_n = a_1 - a_2 + a_3 - a_4 + a_5 -...$ , then:

$s_{2n}$ = $\left\{a_1 - a_2, a_1 - a_2 + a_3 - a_4, a_1 - a_2 + a_3 - a_4 + a_5 - a_6, a_1 - a_2 + a_3 - a_4 + ... + a_{2n} \right\}$

$s_{2n+1}$ = $\left\{ a_1, a_1 - a_2 + a_3, a_1 - a_2 + a_3 - a_4 + a_5, a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + a_7, ...\right\}$

We will show that both sequences converge and converge to the same limit. Then we will show that when shuffled, that sequence will converge. The shuffled sequence will be $s_n$

$\textbf{(i):}$ $s_{2n+1}$ $\leq$ $\left\{a_1, a_1 - a_2 + a_2, a_1 - a_2 + a_2 - a_4 + a_5, ...\right\}$ We are making the additions larger.

= $\left\{a_1, a_1, a_1, ...\right\}$ which is bounded by $2a_1$. converges obviously. This sequence is monotone decreasing (proof by induction). Thus, by the MTC theorem, this sequence converges.

$\textbf{(ii):}$ $s_{2n}$ $\leq$ $\left\{a_1 - a_2, a_1 - a_3 + a_3 - a_4, a_1 - a_3 + a_3 - a_5 + a_5 - a_6, ...\right\}$

= $\left\{a_1 - a_2, a_1 - a_4. a_1 - a_6, a_1 - a_8, ...\right\}$ is clearly bounded by $2a_1$ (since at some point, we are subtracting "$0$"). This sequence is monotone increasing (proof by induction, but to lengthy to put here). Thus, by the MTC theorem, this sequences converges.

$\textbf{Next,}$ We will show both sequences have the same limit. Suppose the limit of $s_{2n}$ is A, limit of of $s_{2n+1}$ is B.

Then the limit of ($s_{2n}$ - $s_{2n+1}$) = $\left\{-a_2, -a_4, -a_6, -a_8,... \right\}$ which converges to $0$.

Thus limit of ($s_{2n}$ - $s_{2n+1}$) = A - B = $0$ implies A = B.

Now, if we $\textbf{shuffle}$ the sequences together, it is clear that we get back $s_n$. More importantly, we now know this sequence converges to A.

Reasoning:

Since $s_{2n}$ converges, $\forall \epsilon > 0, \exists N_1 \in \mathbb{N} | n \geq N_1, |s_{2n} - A| < \epsilon$

Since $s_{2n+1}$ converges, $\forall \epsilon > 0, \exists N_2 \in \mathbb{N} | n \geq N_2, |s_{2n+1} - A| < \epsilon$ (B = A)

If we set N = max$\left\{N_1, N_2\right\}$, then $\forall \epsilon > 0$ when $n \geq N, |s_n - A| < \epsilon$

Since $s_n$ is the sequence of partial sums for the series $$\sum_{n=1}^{\infty} (a_n)*(-1)^{n+1}$$ The series must converge since $s_n$ converges.

• Write $a = \lim_{n\to \infty} s_{2n}$ and $b=\lim_{n\to \infty} s_{2n+1}$. You just need to show $a=b$ to get that the entire sequence $s_n$ converges to $a=b$. For this, just note that $s_{2n+1} = s_{2n} + a_{2n+1}$. – Jeff Oct 6 '16 at 3:49
• I am not sure I follow. You make sense when you devise a way to show that the limits equal each other. I fail to see how two sub-sequences converging to the same limit illustrates the original sequence converging to the same limit. – northcity4 Oct 6 '16 at 15:40
• Because the two subsequences make up the whole sequence when interwoven together. Try proving it rigorously using the definition of a limit: If $\lim_{n\to \infty} s_{2n} = a=\lim_{n\to \infty} s_{2n+1}$ then $\lim_{n\to \infty} s_n = a$. – Jeff Oct 6 '16 at 15:59
• Alright thanks, that makes a lot more sense. I will try to post my solution (using your idea) here at another time. – northcity4 Oct 6 '16 at 19:24
• For (i) you cannot use comparison to prove convergence of sequences. Show that $s_{2n+1}$ is bounded an monotone, as you did for (ii). – Jeff Oct 7 '16 at 14:11