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March 27, 2017

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Convert this series to closed form:

Sn = ln(1 - 1/4) + ln(1 - 1/9) + ln(1 - 1/16) + ... + ln(1 - 1/(n + 1)^2)

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I would guess the first step is this:

e^Sn = (1 - 1/4) * (1 - 1/9) * (1 - 1/16) * ... * (1 - 1/(n + 1)^2)

But how to I conver that to closed form?

  • Math (Series) - ,

    great idea to start, now
    change each of the brackets to a single number, so you would have

    Sn = ln[3/4 * 8/9 * 15/16 * 24/25 * 35/36 + ... ]

    let's start multiplying these
    P1 = 3/4
    P2 = (3/4)(8/9) = 2/3
    P3 = (3/4(8/9)(15/16) = 5/8
    P4 = (3/4(8/9)(15/16)(24/25) = 3/5
    P5 = .... = 7/12

    now take a closer look at
    3/4, 2/3, 5/8, 3/5, 7/12, ...

    or
    3/4, 4/6, 5/8, 6/10, 7/12, ... AHHH

    so Pn = (n+2)/(2n+2)

    Then Sn = ln[(n+2)/(2n+2)]

    or e^(Sn) = (n+2)/(2n+2)

    hope this is what you were looking for, if not, I enjoyed the manipulation anyway.

  • Math (Series) - ,

    Thanks so much Reiny! That's the right answer. I've been struggling with it for a few hours. Here's a slightly different route to the same answer:

    Sn = Π (1 to n) of (1 - 1/(k + 1)^2)
    = Π (1 to n) of (k^2 + 2k)/(k + 1)^2

    Since
    Π of X * Y = Π X * Π Y
    and
    Π (1 to n) of k = n!
    and
    Π (1 to n) of (k + a) = (n + a)!/a!

    Π (1 to n) of (k^2 + 2k)/(k + 1)^2
    = (Π k) * (Π k + 2) * (Π 1/(k + 1)) * (Π 1/(k + 1))
    = n! * (n + 2)!/2 * 1/(n + 1)! * 1/(n + 1)!
    = 1/2 * (n + 2) / (n + 1)

  • Math (Series) - ,

    That's e^Sn... Not Sn...

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