The sum of series with natural logarithm: $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$

Question

Calculate the sum of series:

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

I tried to spread this logarithm, but I'm not seeing any method for this exercise.

Answer 1

Note $$\ln\left(\frac{n(n+2)}{(n+1)^2}\right)=\ln\left(\frac{\frac{n}{n+1}}{\frac{n+1} {n+2}}\right)=\ln\left(\frac{n}{n+1}\right)-\ln\left(\frac{n+1}{n+2}\right)$$ Thus $$\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)=\sum_{n=1}^\infty \left[\ln\left(\frac{n}{n+1}\right)-\ln\left(\frac{n+1}{n+2}\right)\right]$$ This is a telescoping series. Therefore $$\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)=-\ln(2)$$

Answer 2

An overkill. Since holds $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d $$ and this can be proved using the Euler's definition of the Gamma function, we have $$\sum_{n\geq1}\log\left(\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\log\left(\prod_{n\geq0}\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+2\right)}\right)=\log\left(\frac{\Gamma\left(2\right)\Gamma\left(2\right)}{\Gamma\left(1\right)\Gamma\left(3\right)}\right)=\color{red}{\log\left(\frac{1}{2}\right)}.$$

Answer 3

In another, more straight, way: $$\begin{gathered} \sum_{1\leqslant n} \ln \left( n(n+2)\over(n+1)^2 \right) = \ln \left(\prod\limits_{1\leqslant n} {n(n+2)\over(n+1)^2 } \right)\\ = \ln \left({\prod\limits_{1\leqslant n} n \over \prod\limits_{1\leqslant n} (n+1)}\;{\prod\limits_{1\leqslant n} (n+2) \over \prod\limits_{1\leqslant n} (n+1)}\right) = \ln \left({\prod\limits_{1\leqslant n} n \over \prod\limits_{2\leqslant n} n}\;{\prod\limits_{3\leqslant n} n \over \prod\limits_{2\leqslant n} n}\right)\\ = \ln \left(1\cdot\frac12\right) = \ln \left(\frac12\right) \end{gathered}$$

Rewriting the above in more rigorous terms, we have $$\begin{gathered} \sum_{n=1}^q \ln \left(n(n+2)\over(n+1)^2\right) = \ln\left(\prod_{n=1}^q{n(n+2)\over(n+1)^2}\right)\\ = \ln\left({\prod\limits_{n=1}^q n\over\prod\limits_{n=1}^q (n+1)}{\prod\limits_{n=1}^q (n+2)\over\prod\limits_{n=1}^q (n+1)}\right) = \ln\left({\prod\limits_{n=1}^q n\over\prod\limits_{n=2}^{q+1} n}{\prod\limits_{n=3}^{q+2} n\over\prod\limits_{n=2}^{q+1} n}\right)\\ = \ln\left(\frac1{q+1}\;\frac{q+2}2 \right) \end{gathered}$$ and therefore $$ \lim_{q\to\infty}\sum_{n=1}^q\ln\left(n(n+2)\over(n+1)^2\right) = \lim_{q\to\infty}\ln\left(\frac12{q+2\over q+1}\right) = \ln\left(\frac12\lim_{q\to\infty}{q+2\over q+1}\right) = \ln\left(\frac12\right) $$

Answer 4

We can see that $$\frac {n (n+2)}{(n+1)^2} =1-\frac {1}{(n+1)^2} $$ So we have, $$\sum_{n=1}^{\infty} \log \left(1-\frac {1}{(n+1)^2}\right) =\lim_{n \to \infty} \log \left(\left(1-\frac {1}{4}\right)\left(1-\frac {1}{9}\right)\cdots\left(1-\frac {1}{(n+1)^2}\right)\right) =\log \frac {1}{2} $$ Hope it helps.

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For why the infinite product is $\frac {1}{2} $, see here.

Answer 5

Another overkill. Since: $$ \frac{\sin(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)\tag{1} $$ we have: $$ \sum_{n\geq 1}\log\frac{n(n+2)}{(n+1)^2}=\log\prod_{n\geq 2}\left(1-\frac{1}{n^2}\right)=\log\lim_{x\to 1}\frac{\sin(\pi x)}{\pi x(1-x^2)}\stackrel{dH}{=}\color{red}{-\log 2}.\tag{2} $$

Answer 6

\begin{align*}\sum_{n=1}^N \ln\left(\frac{n(n+2)}{(n+1)^2}\right)&= \sum_{n=1}^N \left(\ln n + \ln (n+2) -2\ln(n+1)\right)\\&=\ln 1+\ln(N+2)-\ln2-\ln(N+1)\\&=-\ln2+\ln\frac{N+2}{N+1}\\&\xrightarrow{N\to\infty}-\ln2+\ln1=-\ln2.\end{align*} This is however essentially the same solution as that given by Behrouz, just less clever and explicit about the limit.