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lim( 1/1.3+1/3.5+...1/(2n-1)(2n+1))

where x tends to infinty?

where x tends to infinty?

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- Thread starter succhi
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lim( 1/1.3+1/3.5+...1/(2n-1)(2n+1))

where x tends to infinty?

where x tends to infinty?

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- #2

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Write the first several terms out in that form and see what happens.

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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

- #4

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HallsofIvy said:

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).

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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).

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