# Sum of an Infinite Series

## Homework Statement

Determine the sum of the following series:

$$\sum_{n=1}^{inf} log(1-1/(n+1)^2)$$

Sorry for poor latex, that is supposed to say infinity.

## Homework Equations

How might we turn this into an easier function to deal with?

## The Attempt at a Solution

So far I've only proved convergence of the series. I'm not really sure where to begin. Any help is appreciated. I've ran it through maple and come out with the sum equaling -ln(2).

I thought of trying to work backwards to see if this series was some sort of taylor expansion but I failed at that. I just don't see any elementary techniques. Obviously maple spotted something I haven't lol.

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Maybe it is a geometric series. If so, why not figure out the 1st and 2nd term and the find the r. You are given the first term then use the formula a/ (1-r) where a is the first term and r is the rate.

Okay,

So I worked out the first two terms:

A1 = log(1-1/4) = log(3/4)

A2 = log(1-1/9) = log(8/9)

then r = log(8/9) / log(3/4)

But then checking A3 = log(1-1/16),

A3 does not equal r*A2.

So this is not a geometric series?

Dick
Homework Helper
It's not geometric. Your last hope is that it is a telescoping series. Is it? Write 1-1/(n+1)^2 as a single ratio and factor it up cleverly.

So this is not a geometric series?
Defenitely not.

$$\sum ln\left(1-\frac{1}{(n+1)^2}\right)=\sumln\left(\frac{(n+1)^2-1}{(n+1)^2}\right)=...=$$

$$=\sum ln\keft(\frac{n(n+2)}{(n+1)^2}\left)=...=\sum[ln(n)+ln(n+2)-2ln(n+1)]=$$

$$=\sum[ln(n+2)-ln(n+1)]+\sum [ln(n)-ln(n+1)]$$

Now stuff will cancel out, i already did more than i was supposed to.

Edit: Sorry Dick, i didn't know you were already on it!

Dick