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: $$ \begin{gathered} \sum\limits_{1\, \leqslant \,n} {\ln \left( {\frac{{n\left( {n + 2} \right)}} {{\left( {n + 1} \right)^{\,2} }}} \right)} = \ln \left( {\prod\limits_{1\, \leqslant \,n} {\frac{{n\left( {n + 2} \right)}} {{\left( {n + 1} \right)^{\,2} }}} } \right) = \hfill \\ = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n} n }} {{\prod\limits_{1\, \leqslant \,n} {\left( {n + 1} \right)} }}\;\frac{{\prod\limits_{1\, \leqslant \,n} {\left( {n + 2} \right)} }} {{\prod\limits_{1\, \leqslant \,n} {\left( {n + 1} \right)} }}} \right) = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n} n }} {{\prod\limits_{2\, \leqslant \,n} n }}\;\frac{{\prod\limits_{3\, \leqslant \,n} n }} {{\prod\limits_{2\, \leqslant \,n} n }}} \right) = \hfill \\ = \ln \left( {1\;\frac{1} {2}} \right) = \ln \left( {\frac{1} {2}} \right) \hfill \\ \end{gathered} $$

Rewriting the above in more rigorous terms, we have $$ \begin{gathered} \sum\limits_{1\, \leqslant \,n\, \leqslant \,q} {\ln \left( {\frac{{n\left( {n + 2} \right)}} {{\left( {n + 1} \right)^{\,2} }}} \right)} = \ln \left( {\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\frac{{n\left( {n + 2} \right)}} {{\left( {n + 1} \right)^{\,2} }}} } \right) = \hfill \\ = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} n }} {{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 1} \right)} }}\;\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 2} \right)} }} {{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} {\left( {n + 1} \right)} }}} \right) = \ln \left( {\frac{{\prod\limits_{1\, \leqslant \,n\, \leqslant \,q} n }} {{\prod\limits_{2\, \leqslant \,n\, \leqslant \,q + 1} n }}\;\frac{{\prod\limits_{3\, \leqslant \,n\, \leqslant \,q + 2} n }} {{\prod\limits_{2\, \leqslant \,n\, \leqslant \,q + 1} n }}} \right) = \hfill \\ = \ln \left( {\frac{1} {{q + 1}}\;\frac{{q + 2}} {2}} \right) \hfill \\ \end{gathered} $$ and therefore $$ \mathop {\lim }\limits_{q\, \to \,\infty } \sum\limits_{1\, \leqslant \,n\, \leqslant \,q} {\ln \left( {\frac{{n\left( {n + 2} \right)}} {{\left( {n + 1} \right)^{\,2} }}} \right)} = \mathop {\lim }\limits_{q\, \to \,\infty } \ln \left( {\frac{1} {2}\;\frac{{q + 2}} {{q + 1}}} \right) = \ln \left( {\frac{1} {2}\mathop {\lim }\limits_{q\, \to \,\infty } \;\frac{{q + 2}} {{q + 1}}} \right) = \ln \left( {\frac{1} {2}} \right) $$

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.