# Sum and convergence of series $\sum\limits_{n=2}^{\infty}\ln \left(1-\frac{1}{n^2}\right)$ [duplicate]

How do I calculate the sum of $$\sum _{n=2}^{\infty \:}\ln \left(1-\frac{1}{n^2}\right)$$ and prove that it is a convergent series?

I tried using comparison by choosing $a_n = -\frac{1}{n^2}$ and saying that if this is a convergent series, then my series is also a convergent one, since the $\lim _{n\to \infty }\left(\frac{\left(\ln\left(1-\frac{1}{n^2}\right)\right)}{-\frac{1}{n^2}}\right)$ would be $1$. I'm not sure if this is the correct way of doing this though, since I'm working with positive term series.

## marked as duplicate by vadim123, Did sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 7 '15 at 16:15

• Hint: $$\prod_{n=2}^N\left(1-\frac1{n^2}\right)=\prod_{n=2}^N\frac{n^2-1}{n^2}=\prod_{n=2}^N\frac{(n-1)(n+1)}{n^2}=\frac{N+1}{2N}$$ – Did Nov 7 '15 at 16:03
• Well I know I could always write my sum as $ln\left(n-1\right)+2ln\left(n\right)+ln\left(n+1\right)$ which I think collapses on summation. But I'm still a bit in the dark here on how to prove that it's convergent. – MikhaelM Nov 7 '15 at 16:07
• Why the \: in \sum _{n=2}^{\infty \:}? To make sure that ∞ is not at the right place? – Did Nov 7 '15 at 16:09
Write $\log(1-\frac{1}{n^2})=\log(\frac{n+1}{n})-\log(\frac{n}{n-1})=u_n-u_{n-1}$