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

• succhi

#### succhi

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

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!