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January 31, 2015

January 31, 2015

Posted by **Sean** on Wednesday, March 18, 2009 at 3:54pm.

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

-----------------------------

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) -
**Reiny**, Wednesday, March 18, 2009 at 5:20pmgreat 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) -
**Sean**, Wednesday, March 18, 2009 at 5:47pmThanks 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:

S_{n}= 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) -
**Sean**, Wednesday, March 18, 2009 at 5:48pmThat's e^S

_{n}... Not S_{n}...

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