Convergence of series log(1-1/n^2)

[Felipe Lincoln]

Homework Statement

Find the sum of $\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right)$

Homework Equations

No one.

The Attempt at a Solution

At first I though it as a telescopic serie:
$\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right) =\ln\left(\dfrac{3}{4}\right) + \ln\left(\dfrac{8}{9}\right) + \ln\left(\dfrac{15}{16}\right) + \dots < 0$
But it doesn't look like so.
we know from it's terms that $s_n$ is in the interval of $(L, \ln\left(3/4\right)]$

 

[tnich]

Homework Statement

Find the sum of $\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right)$

Homework Equations

No one.

The Attempt at a Solution

At first I though it as a telescopic serie:
$\sum\limits_{n=2}^{\infty}\ln\left(1-\dfrac{1}{n^2}\right) =\ln\left(\dfrac{3}{4}\right) + \ln\left(\dfrac{8}{9}\right) + \ln\left(\dfrac{15}{16}\right) + \dots < 0$
But it doesn't look like so.
we know from it's terms that $s_n$ is in the interval of $(L, \ln\left(3/4\right)]$

You are right about its being a telescoping series. How can you factor $1-\frac 1 {n^2}$?

 

[Delta2]

Using basic property of logarithms you can find that it is equal to
$$\ln\prod_{n=2}^{\infty} \frac{(n-1)(n+1)}{n^2}$$

So you can work with that product instead and prove that the product is equal to $\frac{1}{2}$ so the limit is $\ln\frac{1}{2}=-\ln2$

I know that usually is not a good idea to work with products but in this case it works..

Now that I see it again, this product is the product of two "telescopic" products $\prod\frac{n-1}{n},\prod\frac{n+1}{n}$, so I guess the original series must be a sum of two telescopic series.