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

  • Thread starter succhi
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  • #1
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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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  • #2
HallsofIvy
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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.
 
  • #3
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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.[/



hey i have tried by this method but this not giving me the right ans could plz solve this briefly 4 me
 
  • #4
mathman
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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.

You have a sign error! Should be 1/(2n-1)(2n+1)= (1/2)/(2n-1) -(1/2)/(2n+1).
 
  • #5
HallsofIvy
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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.[/



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

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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