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.
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)
-----------------------------
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?
3 answers
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)
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)
That's e^Sn... Not Sn...