# Proof of the Alternating Series Approximation Theorem

1. Oct 17, 2013

### Seydlitz

1. The problem statement, all variables and given/known data
Problem taken from Boas Mathematical Methods book, Section 14 page 35.

Prove that if $S=\sum_{n=1}^{\infty} a_n$ is an alternating series with $|a_{n+1}|<|a_n|$, and $\lim_{n \to \infty} a_n=0$, then $|S-(a_1+a_2+...+a_n)|\leq|a_{n+1}|$.

3. The attempt at a solution
Hint: Group the terms in the error as $(a_{n+1}+a_{n+2})+(a_{n+3}+a_{n+4})+...$ to show that the error has the same sign as $a_{n+1}$

Because alternating series, and $|a_{n+1}|<|a_n|$
If $a_{n+1}>0$ and $a_{n+2}<0$, then $(a_{n+1}+a_{n+2})>0$. Then the same thing will happen with the subsequent group for all $n$, and the whole error will have the same sign as $a_{n+1}$. The same can be shown if $a_{n+1}<0$.

Hint: Group them (the error) as $a_{n+1}+(a_{n+2}+a_{n+3})+(a_{n+4}+a_{n+5})+...$
To show that the error has magnitude less than $|a_{n+1}|$.

It is true that $(a_{n+2}+a_{n+3})=(S_{n+3}-S_{n+1})$. It can be shown using induction that the group of the terms can be represented as the partial sum of the whole series as shown above for all$n$.

Hence it is simple to see that because the series is convergent the partial sum will tend to $S$ and the whole group of terms beside $a_{n+1}$ will go to 0. This implies the equation in the theorem.

Alternatively, it can be shown that because of alternating series $(a_{n+2}+a_{n+3})+(a_{n+4}+a_{n+5})+(...)<0$ because of the grouping of the term. The positive odd terms will be smaller than the negative even terms after $a_{n+1}$ and from this we can directly also get the inequality. $|S-S_n|\leq|a_{n+1}|$

Can you guys help me to verify my flow of thoughts and proof?

Thank You

Last edited: Oct 17, 2013
2. Oct 17, 2013

### brmath

You are correct in your statements about the 1st hint, although you don't make it as clear as you could. If $a_{n+1} > 0$ then $a_{n+1}- a_{n+2}$ > 0 and true for all subsequent pairs. So the error term will have a + sign. Same argument if $a_{n+1}$ < 0.

I got a little lost in your presentation after that. The second hint is a good one, and I think you should show that, but I don't think you do; or else I didn't understand your presentation. I'd start something like this: If $a_{n+1}$ >0, what is the sign of $a_{n+2} +a_{n+3}$?.

Once you've proved that step you can say $S - S_{n} = \sum_{k=1}^\infty a_{n+k}$ = $a_{n+1}$ + $\sum_{k=2}^\infty a_{n+k}$. Now what do you know about that last sum?

I don't see any need for induction in this case.

I would like to point out that we do not know that the series is convergent, unless you refer us to a theorem that says it must be. I would say we are on the verge of proving the series is convergent, although what is here is not written quite carefully enough.

3. Oct 17, 2013

### Seydlitz

Ok no problem with the first hint.

Ok I made a mistake about expressing the term as a sum of the series because you can't just take a limit of it and prove the inequality.

Best wishes.