Find Sum of $\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$. Prove that it converges.

Question

Question : For $$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$$
a. Prove it converges
b. Find the sum

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My Try

$ = \sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)\\ = \ln(1 - \frac{1}{4} ) + \ln(1 - \frac{1}{9} ) + \ln ( 1 - \frac{1}{16})\\ = \ln(\frac{3}{4}) + \ln (\frac{8}{9}) + \ln ( \frac{15}{16})\\ = -.287 + -.117 + -.064 = -.50 $

it converges to $-\frac{1}{2}$.

Answer 1

Hint:

Rewrite it as:

$$\sum_{n=2}^{\infty} ln(\frac{n^{2} - 1}{n^{2}}) = \sum_{n=2}^{\infty} ln( \frac{(n-1)(n+1)}{n^{2}}) = \sum_{n=2}^{\infty} [ ln(n-1) + ln(n+1) - 2ln(n)]$$

So we have: $ln(1) + ln(3) - 2ln(2) + ln(2) + ln(4) - 2ln(3) + ...$

Think about this as a telescoping series.

Answer 2

Hint

$$\ln\left(1-\frac1{n^2}\right)=\ln\left(\frac{(n-1)(n+1)}{n^2}\right)=\ln\left(\frac{n-1}{n}\right)-\ln\left(\frac{n}{n+1}\right)=u_n-u_{n+1}$$ and then telescope.

Answer 3

Notice that $$ a_n=\ln\left(1-\frac{1}{n^2}\right)<0 \quad \forall n \ge 2 $$Since the series $$ \sum_{n=1}^\infty b_n:=\sum_{n=1}^\infty\frac{1}{n^2} $$ is convergent, and $$ \lim_{n\to\infty}\frac{-a_n}{b_n}=\lim_{x\to0}\frac{-\ln(1-x)}{x}=1> 0, $$ thanks to the Limit Comparison Test the series $\sum_{n=2}^\infty-a_n=-\sum_{n=2}^\infty a_n$ is convergent, and so is the series $\sum_{n=2}^\infty a_n$.

Furthermore, for every $n\ge 2$ we have \begin{eqnarray} s_n:&=&\sum_{k=2}^n\ln\left(1-\frac{1}{k^2}\right)=\sum_{k=2}^n\left[\ln(k^2-1)-\ln k^2\right]=\sum_{k=2}^n\left[\ln(k-1)+\ln(k+1)-2\ln k\right]\\ &=&\sum_{k=2}^{n-1}\ln k+\sum_{k=3}^{n+1}\ln k-2\sum_{k=2}^n\ln k=-\ln2+\ln n+\ln(n+1)-2\ln n\\ &=&-\ln2+\ln\left(1+\frac1n\right). \end{eqnarray} Since $$ \lim_{n\to\infty}s_n=\lim_{n\to\infty}\left[-\ln2+\ln\left(1+\frac1n\right)\right]=-\ln 2, $$ it follows that $\displaystyle \sum_{n=2}^\infty\ln\left(1-\frac{1}{n^2}\right)=-\ln2$