Homework Help: 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.