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April 3rd, 2007, 10:16 PM   #1
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Calc 2, Comparison Tests, need help using the Calculator

A question is asking me to approximate the first 100 terms of [Riemann sum
(n=1) 1/(n^3)+1]. I'm not sure how to use my calculator in order to figure out the sum of the series.
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April 4th, 2007, 01:01 AM   #2
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What do you call a Riemann sum ? Are you simply trying to evaluate sum(1+1/n^3) ? I've always heard of Riemann sums in a context of Riemann-integration ...
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April 4th, 2007, 02:10 AM   #3
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Here is the question:

Use the sum of the first 100 terms to approximate the sum of the series ∑(n=1, n=∞) 1/(n^3)+1. Estimate the error involves in this approximation.


To start off I concluded that:


the given series is convergent since

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

by the comparison test.


After continuing this problem, I'm up to the step where I have to solve ∑(n=1, n=100) 1/(n^3)+1, but I don't know how to do it. I could plug in each of the 100 terms and add them up, but I know there is a way to do it using the calculator.
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April 4th, 2007, 04:51 AM   #4
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Do you mean 1/(n^3)+1 or 1/((n^3)+1) ? 1/(n^3)+1 diverges by the nth term test.

lim n → ∞ 1/(n³)+1

= lim n → ∞ ((n³) + 1)/(n³) = 1

=diverges.

1/((n³)+1), however, converges by the comparison test with 1/n³ as you say.

Unfortunately, I don't know how to get a calculator to approximate this sort of problem, but I do know how to write programs for my computer to approximate almost anything:

Code:
import java.util.*;
public class approximate
{
 public static void main (String[] args)
 {
  float answer = 0;
  for(int n=1; n<=100; n++)
  {
   answer += 1/(Math.pow(n,3)+1);
  }
  System.out.println(answer);
 }
}
Approximate answer after 100 terms = 0.6864537
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April 10th, 2007, 01:33 PM   #5
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Thank you
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April 10th, 2007, 04:01 PM   #6
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Quote:
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Say, who are you trying to get revenge on, anyway? :P
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