Lim( 1/1.3+1/3.5+ 1/(2n-1)(2n+1))where x tends to infinty?

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Homework Help Overview

The discussion revolves around evaluating the limit of a series involving the terms 1/(2n-1)(2n+1) as n approaches infinity. The subject area includes calculus and series convergence.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of partial fractions to rewrite the term 1/(2n-1)(2n+1) and explore the implications of this form on the limit. There are attempts to derive the first several terms of the series to analyze its behavior.

Discussion Status

Some participants have pointed out potential errors in the application of partial fractions, specifically regarding sign errors. There is an ongoing exploration of the series' terms and their contributions to the limit, with no clear consensus reached yet.

Contextual Notes

Participants express frustration with the method not yielding the expected results, indicating a need for clarification on the approach being taken. There is an emphasis on ensuring the correct application of mathematical techniques.

succhi
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lim( 1/1.3+1/3.5+...1/(2n-1)(2n+1))
where x tends to infinty?
 
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By "partial fractions", 1/(2n-1)(2n+1)= (1/2)/(2n-1)+ (1/2)/(2n+1).

Write the first several terms out in that form and see what happens.
 
HallsofIvy said:
By "partial fractions", 1/(2n-1)(2n+1)= (1/2)/(2n-1)+ (1/2)/(2n+1).

Write the first several terms out in that form and see what happens.[/



hey i have tried by this method but this not giving me the right ans could please solve this briefly 4 me
 
HallsofIvy said:
By "partial fractions", 1/(2n-1)(2n+1)= (1/2)/(2n-1)+ (1/2)/(2n+1).

Write the first several terms out in that form and see what happens.

You have a sign error! Should be 1/(2n-1)(2n+1)= (1/2)/(2n-1) -(1/2)/(2n+1).
 
succhi said:
HallsofIvy said:
By "partial fractions", 1/(2n-1)(2n+1)= (1/2)/(2n-1)+ (1/2)/(2n+1).

Write the first several terms out in that form and see what happens.[/



hey i have tried by this method but this not giving me the right ans could please solve this briefly 4 me
Then what did you get for, say, the first 5 terms?

mathman said:
You have a sign error! Should be 1/(2n-1)(2n+1)= (1/2)/(2n-1) -(1/2)/(2n+1).
Ouch! The fact that one is positive and the other negative is what makes it all work!
 

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