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The sum of the series 1/1.2 - 1/2.3 + 1/...

The sum of the series `1/1.2 - 1/2.3 + 1/3.4 …` upto `infty` is equal to

A

`log_(e) 2 - 1`

B

`log_(e) 2`

C

`log_(e) (4/e)`

D

`2log_(e) 2 `

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • The sum of the series 1/(2!) - 1/(3!) + 1/(4!) - …….. upto infinity is

    A
    `e^((-1)/2)`
    B
    `e^(1/2)`
    C
    `e^(-2)`
    D
    `e^(-1)`

  • The sum of series 1/2! + 1/4! + 1/6! + …….. is

    A
    `(e - 1)^(2)/(2e)`
    B
    `(e^(2) - 1)/(2e)`
    C
    `(e^(2) - 1)/2`
    D
    `(e^(2) - 1)/e`

  • The sum of the series 1 + 1/(4.2!) + 1/(16.4!) + 1/(64.6!) + ....infty is

    A
    `(e + 1)/sqrt(e)`
    B
    `(e - 1)/sqrt(e)`
    C
    `(e + 1)/(2sqrt(e))`
    D
    `(e - 1)/(2sqrt(e))`

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    The sum of series 1 + 2x + 3x^(2) + 4x^(3) + ...... up to infinity when x lies between 0 and 1 i.e., 0 lt x lt 1 is

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