<s>[INST] 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.
 [/INST] 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}
</s>