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log (n +1) - log n = 2[(1)/(2n+1) + ...

`log (n +1) - log n`
` = 2[(1)/(2n+1) + (1)/(3(2n+1)^(3)) + (1)/(5(2n+1)^(5))+...]`

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log_(e) (n+1) - log_(e) (n-1) = 2 [(1)/(n) + (1)/(3) . (1)/(n^(3)) + (1)/(5). (1)/(n^(5)) + ...]

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Knowledge Check

  • If Lt_(n rarr oo) Sigma (log (n+r) - log n)/(n)= 2 (log 2- (1)/(2)) then Lt_(n rarr oo) (1)/(n^(lamda)) [(n+1)^(lamda) (n+2)^(lamda) ..(n+n)^(lamda)]^(1//n) =

    A
    `(4lamda)/(e )`
    B
    `((4)/(3))^(1//lamda)`
    C
    `((4)/(e ))^(lamda)`
    D
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  • Given that lim_(nto oo) sum_(r=1)^(n) (log (r+n)-log n)/(n)=2(log 2-(1)/(2)) , lim_(n to oo) (1)/(n^k)[(n+1)^k(n+2)^k.....(n+n)^k]^(1//n) , is

    A
    `(4k)/(e)`
    B
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    C
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    D
    `((e)/(4))^k`.

  • Let n = 2006! Then (1)/(log_(2)n) + (1)/(log_(3)n) + ... + (1)/(log_(2006)n) =

    A
    2006
    B
    2005
    C
    2005!
    D
    1

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