# Using induction & subsequences to prove the Alternating Series Test

Recall that the Alternating Series Test says that if $a_n\to 0$ and $a_n\geq\ a_{n+1}$ for all $n\geq\ 1$, then the alternating series $\sum\limits_{n=1}^{\infty} (–1)^{n+1}a_n$ converges; i.e., the sequence of partial sums $s_n= a_1–a_2+ a_3–... ± a_n$ converges.

Consider the subsequences ($s_{2n}$) and ($s_{2n-1}$), and show how the Monotone Convergence Theorem proves the Alternating Series Test.

My work:

Since we need to use the MCT, I was thinking of first showing that the subsequences ($s_{2n}$) and ($s_{2n-1}$) are bounded and monotone. So for the even subsequences I wanted to show they are increasing & bounded above while the odds are decreasing & bounded below.

However, I am having some trouble with this. So, I am not sure if I am proving by induction on $n$ that $s_{2n}\leq\ s_{2n-1}$ or instead that $s_{2n}\leq\ s_1$ and $s_2\leq\ s_{2n-1}$.

Also, to show the evens are increasing, we would need to show $s_{2n+2}-s_{2n}\geq\ 0$, but would I need to first show the other inductions? I am not sure how to show rigorously that the evens are increasing while the odds are decreasing.

Update

Does this work.

I will show by induction on $n$ that $s_{2n}\leq\ s_1$.

For the base case of $n=1$ we have that $a_1-a_2<a_1$ so it is true. Then we assume $s_{2n}\leq\ s_1$.

Then, $s_{2(n+1)}=s_{2n+2}= s_1-a_{2n+2}\leq\ s_1-a_{2n+2}<a_1$. So this shows bounded above.

How can I show increasing in this case?

How can I show that $s_{2n+2}-s_{2n}\geq\ 0$?

I need help with the odd sequences part of the proof. I am not too sure how to show they are bounded below by $s_2$ because I tried induction and it did not work out.

Using induction you can easily prove that $s_{2k+1} \geq s_{2k+3}$ for any $k \geq 0$ and that it also bounded below by zero. Therefore by the Monotone Convergence Theorem the sequence $(s_{2n+1})$ has a limit denoted $l_1 \geq 0$
On the other hand, using an induction argument you can show that $s_{2k} \leq s_{2k+2}$ for any $k \geq 1$, so the sequence $(s_{2n})$ has a limit $l_2$ (Careful as $l_2$ might be $\infty$ apriori).
Finally, using the fact that $a_n\to 0$, conclude that $l_1=l_2$.
• I am not sure how you are doing that first induction. Aren’t we showing that the odd sequences are bounded below by say $s_2$ and the evens are bounded above by say $s_1$? – Sam Oct 22 '17 at 17:23
• You have that $s_{2k+2}=s_{2k}+a_{2k}-a_{2k+1}$, and then use $a_{2k} \geq a_{2k+1}$ – C.Niculescu Oct 22 '17 at 22:55
• Right, I got it, thank you. Do you know how we can split up $s_{2n-1}$? – Sam Oct 23 '17 at 0:25