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

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The limit of the series lim(1/(2n-1)(2n+1)) as n approaches infinity involves using partial fractions. The correct decomposition is 1/(2n-1)(2n+1) = (1/2)/(2n-1) - (1/2)/(2n+1). This sign difference is crucial for the series to converge properly. Participants emphasize the importance of writing out the first several terms to observe the behavior of the series. The discussion highlights the necessity of careful algebraic manipulation to achieve the correct result.
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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