for $f(x)=\ln({1-\frac{1}{x^2})}$ find $\lim_{n\to\infty}{f(1)+f(2)+...+f(n)}$

Question

Let $f:(1, \infty)\to \Bbb R$ with $f(x)=\ln\left({1-\frac{1}{x^2}}\right)$

Find $$\lim\limits_{n\to\infty}\left[f(1)+f(2)+...+f(n)\right]$$

What I have done so far is

$$\begin{align} S_n&=\sum_{k=2}^n{f(k)} \\ &= \ln\left(\prod_{k=2}^n \left(1-\frac{1}{k^2}\right)\right)\\ &= \ln\left(\prod_{k=2}^n\left(\frac{k-1}{k^2}\right)\prod_{k=2}^n\left(\frac{k+1}{k^2}\right)\right)\\ &=\ln\left(\prod_{k=2}^n\left(\frac{1}{k}-\frac{1}{k^2}\right)\prod_{k=2}^n\left(\frac{1}{k}+\frac{1}{k^2}\right)\right) \end{align}$$

Answer 1

Since $f(n)=\ln \frac{(n-1)(n+1)}{n^2}$ for all $n\ge 2$, $$ \sum_{k=2}^n f(n)=\ln \frac{1\cdot 3}{2\cdot 2}\cdot \frac{2\cdot 4}{3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\cdot \cdots \cdot \frac{(n-1)(n+1)}{n\cdot n} =\ln \frac{n+1}{2n}. $$ Therefore, $$ \lim_{n\to\infty}\sum_{k=2}^n f(n)= -\ln 2. $$ (I excluded $f(1)$, because it is not defined.)