This question already has an answer here:

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 Nov 7 '15 at 16:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 4
    $\begingroup$ Hint: $$\prod_{n=2}^N\left(1-\frac1{n^2}\right)=\prod_{n=2}^N\frac{n^2-1}{n^2}=\prod_{n=2}^N\frac{(n-1)(n+1)}{n^2}=\frac{N+1}{2N}$$ $\endgroup$ – Did Nov 7 '15 at 16:03
  • $\begingroup$ 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. $\endgroup$ – MikhaelM Nov 7 '15 at 16:07
  • $\begingroup$ Why the \: in \sum _{n=2}^{\infty \:}? To make sure that is not at the right place? $\endgroup$ – Did Nov 7 '15 at 16:09

Hint: \begin{align} & \log\left(1 - \frac{1}{n^2}\right) = \log(n^2 - 1) - \log(n^2) = [\log(n + 1) - \log n] - [\log(n) - \log(n - 1)] \end{align} Now consider the partial sum.


Write $\log(1-\frac{1}{n^2})=\log(\frac{n+1}{n})-\log(\frac{n}{n-1})=u_n-u_{n-1}$


Not the answer you're looking for? Browse other questions tagged or ask your own question.