Prove that $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ converges and find its sum

Solution 1:

The series $$\sum_{n=1}^{\infty}\ln\left(\frac{n(n+2)}{(n+1)^2}\right)$$ converges, using the telescoping technique: $$\begin{align} \sum_{n=1}^{\infty}\ln\left(\frac{n(n+2)}{(n+1)^2}\right)&=\left[\ln(1)+\ln(3)-2\ln(2)\right]+\left[\ln(2)+\ln(4)-2\ln(3)\right]+\left[\ln(3)+\ln(5)-2\ln(4)\right]+\cdots \end{align}$$

Note that the partial sums are $$\begin{align} S_1&=\ln(1)+\ln(3)-2\ln(2)\\ S_2&=\ln(1)+\ln(3)-2\ln(2)+\ln(2)+\ln(4)-2\ln(3)\\ &=\ln(1)-\ln(2)-\ln(3)+\ln(4)\\ S_3&=S_2+\ln(3)+\ln(5)-2\ln(4)\\ &=\ln(1)-\ln(2)-\ln(4)+\ln(5) \end{align}$$ And in general, $S_n=\ln(1)-\ln(2)-\ln(n+1)+\ln(n+2)=\ln\left(\frac{n+2}{2(n+1)}\right)$ which approaches $\ln(1/2)$.

Solution 2:

The series can be turned into telescopic as follow \begin{align} S_m=\sum_{n=1}^{m}\ln\frac{n(n+2)}{(n+1)^2}&=\sum_{n=1}^m(\ln{(n+2)}+\ln n-2\ln{(n+1)}) \\ &=\sum_{n=1}^m((\ln{(n+2)}-\ln(n+1))-(\ln{(n+1)}-\ln n))) \\ &=\ln{(m+2)}-\ln(m+1)-(\ln{(2)}-\ln(1)) \end{align} So $$ \lim_{m\to\infty}S_m=\lim_{m\to\infty}\left(\ln\frac{m+2}{m+1}-\ln2\right)=-\ln2 $$

Solution 3:

$$ \begin{align} \sum_{k=1}^\infty\log\left(\frac{k(k+2)}{(k+1)^2}\right) &=\log\left(\prod_{k=1}^\infty\frac{k(k+2)}{(k+1)^2}\right)\\ &=\lim_{n\to\infty}\log\left(\prod_{k=1}^n\frac{k(k+2)}{(k+1)^2}\right)\\ &=\lim_{n\to\infty}\log\left(\frac{\Gamma(n+1)}{\Gamma(1)}\frac{\Gamma(n+3)}{\Gamma(3)}\frac{\Gamma(2)^2}{\Gamma(n+2)^2}\right)\\ &=-\log(2)+\lim_{n\to\infty}\log\left(\frac{\Gamma(n+1)\Gamma(n+3)}{\Gamma(n+2)^2}\right)\\[6pt] &=-\log(2) \end{align} $$ The last limit is a consequence of Gautschi's Inequality.

Another Approach

Prove by induction that $$ \prod_{k=1}^n\frac{k(k+2)}{(k+1)^2}=\frac12\frac{n+2}{n+1} $$