Solving Series w/ Remainder Estimate & Integral Test

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SUMMARY

The discussion focuses on using the Remainder Estimate for the Integral Test to determine the sum of the series ∑n=1 (1/n3) with an accuracy of three decimal places. Participants clarify that to achieve this accuracy, the error must be less than 0.0005 to avoid rounding issues that could affect the third decimal place. The consensus is that using an error threshold of 0.0005 is appropriate for this calculation.

PREREQUISITES
  • Understanding of the Integral Test for convergence of series
  • Familiarity with Remainder Estimates in calculus
  • Knowledge of series summation techniques
  • Basic proficiency in error analysis and rounding principles
NEXT STEPS
  • Study the Remainder Estimate for the Integral Test in detail
  • Learn about error analysis in numerical methods
  • Explore convergence tests for series, particularly p-series
  • Practice calculating series sums with specified accuracy thresholds
USEFUL FOR

Students in calculus courses, educators teaching series convergence, and anyone interested in numerical analysis and error estimation techniques.

skyturnred
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It asks "use remainder estimate for integral test" to find series accurate to 3 dec?

Homework Statement



It says "Use the Remainder Estimate for the Integral Test to find the sum of the following series to three decimal places of accuracy."

\sum^{\infty}_{n=1} \frac{1}{n^{3}}

Homework Equations





The Attempt at a Solution



OK, so my question is as follows. If it says "to 3 decimal places of accuracy," what am I finding exactly? Am I finding the sum for error<0.001? or is it error<0.0005? I ask this because an error of 0.0005 could change the round and change the third decimal place.

As for the rest of the question, I know how to do it.

Thanks!
 
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skyturnred said:

Homework Statement



It says "Use the Remainder Estimate for the Integral Test to find the sum of the following series to three decimal places of accuracy."

\sum^{\infty}_{n=1} \frac{1}{n^{3}}

Homework Equations





The Attempt at a Solution



OK, so my question is as follows. If it says "to 3 decimal places of accuracy," what am I finding exactly? Am I finding the sum for error<0.001? or is it error<0.0005? I ask this because an error of 0.0005 could change the round and change the third decimal place.

As for the rest of the question, I know how to do it.

Thanks!

i have always used < 0.0005, just because of the possibility of "rounding error" changing the 3rd decimal place.
 


Thanks! Thats what Ill do!
 

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