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(1/1.3 + 1/3.5 + ... + 1/(2n-1)(2n+1)) as n tends to infinity can be evaluated 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 convergence of the series. The first several terms should be written out to observe the behavior of the series as n increases.

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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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