Convergence of ln(1 - 1/n^2) Series: Proving -ln 2 as the Limit

[merced]

Show that Σ$$_{n=1}^\infty ln(1 -1/n^2) = -ln 2$$

I'm not sure how to do this. Should I use telescopic sums or should I make a function $$y = ln(1 - 1/n^2)$$? Is it possible to use telescopic sums here?

 

[merced]

When I make $$y = ln(1 - 1/n^2)$$..then the limit of y = 0 which yields no conclusion about the series.

 

[quasar987]

The sum cannot start at n=1 because then the first term blows up. I get the correct answer if n starts at 2. Two hints:

1) Yes it's a telescoping series problem.

2) ln(a+b) is not a friendly expressiom but ln(a*b) is because ln(a*b)=lna + lnb. And notice that $$1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{(n-1)(n+1)}{n^2}$$

 

[quasar987]

Incidentally, this problem proves indirectly that

$$\prod_{i=2}^{\infty}\frac{i^2-1}{i^2}=\left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)\left(\frac{24}{25}\right)...=\frac{1}{2}$$

Nice!

 

[merced]

Thank you!