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

• Felipe Lincoln

Gold Member

## Homework Statement

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

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)]##

## Homework Statement

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

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}##?

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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.

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