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.


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.

  • 1
    $\begingroup$ 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 $s_{2n+1} = s_{2n} + a_{2n+1}$. $\endgroup$ – Jeff Oct 6 '16 at 3:49
  • $\begingroup$ 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. $\endgroup$ – northcity4 Oct 6 '16 at 15:40
  • $\begingroup$ 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 $\lim_{n\to \infty} s_n = a$. $\endgroup$ – Jeff Oct 6 '16 at 15:59
  • $\begingroup$ Alright thanks, that makes a lot more sense. I will try to post my solution (using your idea) here at another time. $\endgroup$ – northcity4 Oct 6 '16 at 19:24
  • $\begingroup$ 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). $\endgroup$ – Jeff Oct 7 '16 at 14:11

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