succhi
- 4
- 0
lim( 1/1.3+1/3.5+...1/(2n-1)(2n+1))
where x tends to infinty?
where x tends to infinty?
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.
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.
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.
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.
succhi said:Then what did you get for, say, the first 5 terms?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
Ouch! The fact that one is positive and the other negative is what makes it all work!mathman said:You have a sign error! Should be 1/(2n-1)(2n+1)= (1/2)/(2n-1) -(1/2)/(2n+1).