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?

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.

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)

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

To convert the series to closed form, we can make use of the properties of logarithms and exponentiation.

Let's start by re-expressing the equation you provided:

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

We can observe that each term in the product corresponds to:

(1 - 1/(k + 1)^2)

Let's simplify this term further:

(1 - 1/(k + 1)^2) = ((k + 1)^2 - 1) / (k + 1)^2

We can rewrite this expression as:

((k + 1)^2 - 1) = k^2 + 2k

Now, our equation becomes:

e^Sn = [(1^2 + 2 * 1) / 1^2] * [(2^2 + 2 * 2) / 2^2] * [(3^2 + 2 * 3) / 3^2] * ... * [(n^2 + 2 * n) / n^2]

Simplifying further, we have:

e^Sn = (1^2 + 2 * 1) * (2^2 + 2 * 2) * (3^2 + 2 * 3) * ... * (n^2 + 2 * n) / (1^2 * 2^2 * 3^2 * ... * n^2)

Now, if we take the natural logarithm (ln) of both sides, we get:

Sn = ln[(1^2 + 2 * 1) * (2^2 + 2 * 2) * (3^2 + 2 * 3) * ... * (n^2 + 2 * n) / (1^2 * 2^2 * 3^2 * ... * n^2)]

This provides the closed form for the given series.

Note that for large values of n, calculating the closed form directly might be computationally expensive due to the presence of factorials in the denominator. In such cases, it might be more practical to approximate the sum using numerical methods or software.