# Proof of the Alternating Series Approximation Theorem

Seydlitz

## Homework Statement

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}|##.

## 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

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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.

Seydlitz
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.

Ok no problem with the first hint.

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.

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.

I just realized that the alternative method makes more sense because if ##a_{n+1}>0##, certainly the rest of the sum would be negative## and with it I can get the inequality in the theorem. I think the hint is very useful if you think it in terms of the sign.

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.

I think the book also assumes that the series meets the alternating series for convergence, and hence it's convergence.

Thanks

brmath
I think the book also assumes that the series meets the alternating series for convergence, and hence it's convergence.

Thanks

Yes, it is true that alternating series which has terms going to zero will converge; and this problem nearly proves it. The only problem with this as a proof is that you can't assume an S if there may not be one. But that can be worked around.

Seydlitz
Yes, it is true that alternating series which has terms going to zero will converge; and this problem nearly proves it. The only problem with this as a proof is that you can't assume an S if there may not be one. But that can be worked around.

Can we just assume there's always S? How do you suggest we go around it?

brmath
If it doesn't converge, there wouldn't be a finite S. Better practice is to write it as
##\sum_{k=1}^{\infty} - \sum_{k=1}^n \le |a_{n+1}|##,
or if you want to save writing begin by defining S = ##\sum_{k=1}^{\infty}##.

In this case the distinction is trivial because you have that strong inequality, which implies that there must be convergence. However, I've had a lot of problems in the past with writing things sloppily and then having trouble, so I tend to be careful.

Seydlitz
If it doesn't converge, there wouldn't be a finite S. Better practice is to write it as
##\sum_{k=1}^{\infty} - \sum_{k=1}^n \le |a_{n+1}|##,
or if you want to save writing begin by defining S = ##\sum_{k=1}^{\infty}##.

In this case the distinction is trivial because you have that strong inequality, which implies that there must be convergence. However, I've had a lot of problems in the past with writing things sloppily and then having trouble, so I tend to be careful.

Ok I understand that it really needs to be very clear.

You're summing ##a_n## right?

brmath
Yes I am summing ##a_n##, but my Latex fingers were tired. So now I am the one not being completely clear.

Best wishes.