Infinite sum including logarithm I would like to calculate the following sum:

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

I do know that it converges but I have gone that far:
\begin{align}
& \sum_{n=1}^{\infty}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right) \Longleftrightarrow \sum_{n=1}^\infty \ln \left( \frac{(n+1)^2}{n(n+2)} \right) \\[10pt]
= {} & \ln \left( \frac{4}{3} \right)+\ln \left( \frac{9}{8} \right)+\ln \left( \frac{16}{15} \right)+\cdots+\ln \left( \frac{n}{n-1} \right)\\[10pt]
= {} & \ln \left( \frac{4}{3}\frac{9}{8}\frac{16}{15} \cdots \frac{n}{(n-1)} \right)=\ln (n)
\end{align}
which diverges as $n\to \infty.$
It looked like  telescoping in the beginning but now I am confused. Where have I gone wrong?
Thanks.
 A: Rewrite the general term as $\ln\dfrac{(n+1)^2}{n(n+2)}$, use the functional property of logs and you'll obtain a telescoping product for partial sums:
\begin{align}\sum_{k=1}^{n}\ln \frac{(k+1)^2}{k(k+2)}&=\ln\frac{2^2}{1\cdot 3}+\ln\frac{3^2}{2\cdot 4}+\ln\frac{4^2}{3\cdot 5}+\dotsm\dotsm\dotsm\\&\phantom{=}+\ln\frac{(n-1)^2}{(n-2)n}+\ln\frac{n^2}{(n-1)(n+1)}+\ln\frac{(n+1)^2}{n(n+2)}\\
&=\ln\frac{2^{\not2}\cdot\not 3^2\cdot4^2\dotsm\dotsm\dotsm\dotsm(n-1)^2\not n^2(n+1)^2\hspace{3em}}{1\cdot \not3\cdot\not2\cdot 4\cdot\not 3\cdot 5\dotsm(n-2)\not n(n-1)(n+1)\not n(n+2)}\\
&=\log\frac{2(n+1)}{n+2}\to \ln 2.\end{align}
A: $\begin{array}\\
\sum_{n=1}^{m}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right)
&=\sum_{n=1}^{m}\ln \left( \frac{(n+1)^2}{n(n+2)} \right)\\
&=\sum_{n=1}^{m}( (2\ln(n+1)-\ln(n)-\ln(n+2))\\
&=2\sum_{n=1}^{m}\ln(n+1)-\sum_{n=1}^{m}\ln(n)-\sum_{n=1}^{m}\ln(n+2)\\
&=2\sum_{n=2}^{m+1}\ln(n)-\sum_{n=1}^{m}\ln(n)-\sum_{n=3}^{m+2}\ln(n)\\
&=2(\ln(2)+\ln(m+1)+\sum_{n=3}^{m}\ln(n))-(\ln(1)+\ln(2)+\sum_{n=3}^{m}\ln(n))-(\ln(m+1)+\ln(m+2)+\sum_{n=3}^{m}\ln(n))\\
&=2(\ln(2)+\ln(m+1))-(\ln(2))-(\ln(m+1)+\ln(m+2))\\
&=\ln(2)+\ln(m+1)-\ln(m+2)\\
&=\ln(2)+\ln(1-\frac{1}{m+2})\\
&\to \ln(2)
\qquad\text{since } \ln(1-\frac{1}{m+2}) \to 0
\text{ as } m \to \infty\\
\end{array}
$
