# Computig the series $\sum\limits_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$

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.

\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*}

• Just an extra, it is important to note that for n to infinity, the infinite terms, when putting the log terms together, approach log1, which is zero. It is always important to check what those infinite terms do. Sometimes they do contribute to an answer. The arctan is a classical example. – imranfat Aug 25 '13 at 22:39

$$\sum_{n=2}^{\infty} \ln \left( 1 - \frac{1}{n^2} \right) = \ln \left( \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2} \right) \right)$$

$$= \ln \left( \prod_{n=2}^{\infty} \frac{n-1}{n}\frac{n+1}{n} \right)$$

$$\ln \left( \frac{1}{2} \times \frac{3}{2} \times \frac{2}{3} \times \frac{4}{3} \times \frac{3}{4} \times ... \right)$$

$$= \ln \left( \frac{1}{2} \right) = -\ln 2$$

• neat -- as expected you can also reduce it to a telescoping product. +1 – oldrinb Aug 25 '13 at 22:39

An overkill is as follows $$\pi^2\frac{\sin x} {x(\pi^2-x^2)} =\prod_{n=2}^{\infty} \left(1-\frac {x^2}{n^2\pi^2}\right)$$ Taking limits as $x\to\pi$ and then applying logarithm we get the desired result.

• Sorry for the question , but where does the product series equality come from? – gimusi Feb 1 '18 at 8:29
• @gimusi : it's the infinite product for $\sin x$ given by $$\sin x=x\prod_{n=1}^{\infty} \left(1-\dfrac{x^2}{n^2\pi^2}\right)$$ – Paramanand Singh Feb 1 '18 at 10:50
• Nice trick. I was wondering how to derive this product series, could you please indicate to me some reference? Thanks – gimusi Feb 1 '18 at 11:01
• – Paramanand Singh Feb 1 '18 at 11:21
• Thanks a lot for the reference! – gimusi Feb 1 '18 at 11:34

## By telescoping

setting $u_n =\ln\left(\frac{n}{n-1}\right)$,we have $$\ln\left(1-\frac{1}{n^2}\right)=\ln\left(1+\frac{1}{n}\right)+\ln\left(1-\frac{1}{n}\right) =\ln\left(\frac{n+1}{n}\right)-\ln\left(\frac{n}{n-1}\right)$$that is $$\color{blue}{\ln\left(1-\frac{1}{n^2}\right)=u_{n+1}-u_n}$$

Whence, $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)= \sum_{n=2}^\infty (u_{n+1}-u_n) = \lim_{n\to\infty}u_n-u_2 = -u_2 =-\ln 2.$$