So I have this problem for midterm reviews:
$$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)=\text{ ?}$$
I know that you can find the series form of a natural log, as shown here:
$$\ln\left(1-\frac{1}{n^2}\right)=-\sum_{k=2}^\infty \left(\frac{1}{n^{4k}}\right)\left(\frac{1}{2k}\right) $$
But the above doesn't seem to help very much since it results in two summation notations mushed together. Is there a somewhat nontedious way to go about this? Thanks! All help appreciated.
Best Answer
$$\begin{align*}\log(1-1/n^2)&=\log\left(\frac{n^2-1}{n^2}\right)\\&=\log(n^2-1)-\log(n^2)\\&=\log[(n+1)(n-1)]-2\log(n)\\&=\log(n+1)+\log(n-1)-2\log(n)\end{align*}$$... and our series telescopes. The only term that survives is $-\log 2$.$$\begin{align*}\sum_{n=2}^\infty\log(1-1/n^2)&=\sum_{n=2}^\infty\left(\log(n+1)+\log(n-1)-2\log (n)\right)\\&=\log3+\log1\color{red}{-2\log2}+\log4+\color{red}{\log2}-2\log3+\dots\\&=-\log 2\end{align*}$$