Alternating Series Approximation - Please help

In summary, the manual's logic 1/(n+1)^3 < 1/1000 will not yield the right result. Alternating series remainder theorem: |S-Sn| =|Rn|< or = to an+1 allows for an appropriate approximation to the sum without knowing the actual sum.
  • #1
bcjochim07
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1. Homework Statement

Determine the number of terms required to approximate the sum of the series with an error of less than .001

Sum ((-1)^(n+1))/(n^3) from n=1 to infinity

2. Homework Equations



3. The Attempt at a Solution

I guess this is what you do:

1/(n+1)^3 < 1/1000

and solving you get n+1 > 10 so the answer is 10 terms

But that doesn't quite make sense to me, and I'm not sure why.

Alternating series remainder theorem:

|S-Sn| =|Rn|< or = to an+1

Could someone please explain this to me?
 
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  • #2
Using 1/(n+1)^3 < 1/1000 will not yield the right result. See the below logic.

An alternating series alternates between positive and negative terms, correct? If you look at an n-term approximation of the sum of an alternating series, you expect that n-term approximation to alternate between "overshooting" and "undershooting" the total (infinite term) sum, with increasing convergence toward the actual infinite-term summation. Since adding a subsequent term to the summation always takes the sum from "undershooting" to "overshooting" or vice versa, the magnitude of the n+1th term of the sequence is always greater than the error between the n-term sum and the infinite term sum. This is what your remainder theorem says, in words.

The nice part about the theorem is that it gives a way to determine an appropriate approximation without knowing the actual sum. You're given an Rn=0.001, now simply find the smallest a_{n+1} satisfying a_{n+1}>=1/1000. Note if you were to (for example) choose the second smallest a_{n+1}, your error term S-Sn would yield too large an error.
 
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  • #3
Yeah, I see what you're saying and that's why I'm confused. The logic 1/(n+1)^3 < 1/1000 is what my solutions manual says. Then they go on to say that n+1 > 10 and that you need 10 terms.

What would you say that the correct answer is?
 
  • #4
Solutions manuals aren't right all the time. Indeed, the difference between the solutions manual and the theorem is 1 term (9 terms is the right answer), but the requirement in the solutions manual is tighter than needed and fails to take into account the nuances of alternating series.
 
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  • #5
Actually, I've been doing some thinking, and I think that the manual is right.

According the manual's logic, the an+1 term is less than 1/1000, and according to the theorem, our error (the remainder) is less than or equal to the an+1 term. So if an+1 is less than 1/1000 then the error must also be less than 1/1000. So I think that if you find n so that 1/(n+1)^3 = 1/1000 (n=9) and then go up to the next integer so that you satisfy the inequality, 10, that is the correct # of terms.
 
  • #6
It's definitely a subtle point, but my answer is vindicated by actually computing the sequence. The sequence converges to 0.9015426... (and so on), and the 9-term summation is 0.90207, which is clearly within the error margin of 1/1000.

So if an+1 is less than 1/1000 then the error must also be less than 1/1000.
Correct, but if a_{n+1} is NOT less than 1/1000, the error isn't necessarily greater than 1/1000 because Sn+a_{n+1}-S is of the opposite sign as Sn-S, so some fraction of a_{n+1} added to Sn is equal to S. Let's say a_{n+1}=x+y, where x is the aforementioned fraction, then Sn+x=S. It's evident that the chunk "y" is the only part of a_{n+1} that actually contributes to summation error since Sn+x=S, and therefore Sn+x+y=S+y. Your solutions manual assumes that all of a_{n+1} contributes to error, which is untrue as I have attempted to explain.

If this argument fails, it's ASCII art time :P
 
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  • #7
Wow. I think I understand your argument. This is some pretty deep thinking. I really appreciate your help. Thanks!
 

What is alternating series approximation?

Alternating series approximation is a method used in mathematics to estimate the value of a series by using a simpler series that alternates between positive and negative terms.

How is alternating series approximation used?

Alternating series approximation is used by replacing the original series with a simpler alternating series that is easier to calculate. The more terms of the alternating series that are included, the closer the approximation will be to the actual value of the original series.

What is the alternating series test?

The alternating series test is a method used to determine whether an alternating series converges or diverges. It states that if a series alternates between positive and negative terms, and the absolute value of each term decreases as the series progresses, then the series will converge.

What is the error bound for alternating series approximation?

The error bound for alternating series approximation is the difference between the actual value of the original series and the value calculated using the alternating series. It can be calculated using the remainder term of the alternating series, which depends on the number of terms included in the approximation.

What are some real-world applications of alternating series approximation?

Alternating series approximation can be used in a variety of real-world applications, such as calculating the value of financial investments, estimating the value of physical quantities in physics, and approximating the solutions to differential equations in engineering.

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