Examine convergence of the series by definition. I have a problem with this series.
$$\sum_{n=1}^\infty \ln \left (1-\frac{2}{n(n+1)}\right ).$$
I need to examine if the series converges by definition. I know that if $n=1$ we have $-\infty$. Is this sufficient to prove that this one diverges?
 A: Directly applying the definition:
$$
\begin{aligned}
\sum_{n=2}^\infty \ln\left( 1-\frac{2}{n(n+1)}\right) &= \lim_{k\to\infty} \sum_{n=2}^k \ln\left( 1-\frac{2}{n(n+1)}\right)= \\
&= \lim_{k\to\infty} \ln \prod_{n=2}^k \left( 1-\frac{2}{n(n+1)}\right)= \\
&= \lim_{k\to\infty} \ln \prod_{n=2}^k \frac{n^2+n-2}{n(n+1)}= \\
&= \lim_{k\to\infty} \ln \prod_{n=2}^k \frac{(n-1)(n+2)}{n(n+1)}= \\
&= \lim_{k\to\infty} \ln \prod_{n=2}^k \frac{n-1}{n+1}\prod_{n=2}^k \frac{ n+2}{n}= \\
&= \lim_{k\to\infty} \ln \prod_{n=2}^k \frac{n-1}{n+1}\prod_{n=3}^{k+1} \frac{ n+1}{n-1}= \\
&= \lim_{k\to\infty} \ln \frac{1}{3}\frac{ k+2}{k}
\end{aligned}
$$
Can you take it from here?
A: The series can not start at $n=1$ because $\text{ln}(1-\frac{2}{1\times2})=\text{ln}(0)$ which does not exist. You can start with $n=2$. Also $\text{ln}(1-\frac{2}{n(n+1)})=\text{ln}\big(\frac{(n+2)(n-1)}{n(n+1)}\big)$.
To prove that the series converges by definition is saying that  $S_{N}=\sum_{n=2}^{N}\text{ln}\big(\frac{(n+2)(n-1)}{n(n+1)}\big)$ converges:
$$\sum_{n=2}^{N}\text{ln}\big(\frac{(n+2)(n-1)}{n(n+1)}\big)=\text{ln}\big(\frac{4}{2\times 3}\big)+\text{ln}\big(\frac{5\times2}{3\times 4}\big)+\text{ln}\big(\frac{6\times3}{4\times 5}\big)+...+\text{ln}\big(\frac{(N+2)\times(N-1)}{N\times (N+1)}\big)=$$
$$=\text{ln}\Big(\frac{4\times5\times 2\times6\times3\times...\times(N+2)\times(N-1)}{2\times3\times3\times4\times4\times5\times...\times N \times(N+1)} \Big)=\text{ln}\Big(\frac{N+2}{3N}\Big)$$.
Finally when $N\rightarrow \infty$ then $\text{ln}\Big(\frac{N+2}{3N}\Big)\rightarrow\text{ln}\big(\frac{1}{3}\big)$.
