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

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.

## marked as duplicate by Martin Sleziak, Arnaud D., zhoraster, martini, C. FalconJan 8 '17 at 0:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – Martin Sleziak Jan 5 '17 at 7:35
• I am amazed by the number of views and votes here. It doesn't make any sense. – Zaid Alyafeai Jan 5 '17 at 8:01
• @ZaidAlyafeai This usually happens when a question appears in network-wide hot questions list. And this has to do with the fact the the OP did not include the series in the title, see meta: Do questions with LaTeX in titles appear in Hot Questions? – Martin Sleziak Jan 5 '17 at 9:36
• @MartinSleziak, interesting, I never knew that. But still the question has to be interesting to show up in the hot question list? I don't see anything special here ? – Zaid Alyafeai Jan 5 '17 at 9:48

## 6 Answers

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

• Maybe worth pointing out that $\ln\left(\frac{n(n+2)}{(n+1)^2}\right)=\ln\left(\frac{\frac{n}{n+1}}{\frac{n+1} {n+2}}\right)$ only for $n \neq -1, -2$, but since the summation starts at $1$ that's ok. – Chris Bouchard Jan 5 '17 at 14:59

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 $$\sum_{n\geq1}\log\left(\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\right)=\log\left(\prod_{n\geq0}\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+2\right)}\right)=\log\left(\frac{\Gamma\left(2\right)\Gamma\left(2\right)}{\Gamma\left(1\right)\Gamma\left(3\right)}\right)=\color{red}{\log\left(\frac{1}{2}\right)}.$$

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 $$\mathop {\lim }\limits_{q\, \to \,\infty } \sum\limits_{1\, \leqslant \,n\, \leqslant \,q} {\ln \left( {\frac{{n\left( {n + 2} \right)}} {{\left( {n + 1} \right)^{\,2} }}} \right)} = \mathop {\lim }\limits_{q\, \to \,\infty } \ln \left( {\frac{1} {2}\;\frac{{q + 2}} {{q + 1}}} \right) = \ln \left( {\frac{1} {2}\mathop {\lim }\limits_{q\, \to \,\infty } \;\frac{{q + 2}} {{q + 1}}} \right) = \ln \left( {\frac{1} {2}} \right)$$

• For the sake of the beginner... do note that the argument doesn't make sense literally as written. It is a quick and efficient argument for those with the experience to recognize that, in this particular problem the tails don't pose an issue and everything 'just works'. Those without that experience should take the further steps to turn it into a rigorous argument. – Hurkyl Jan 4 '17 at 16:52
• @Hurkyl: If I understand your critics, you are hinting to the problem of preserving convergence when splitting the product, which is in fact a fully right concern, that in this case I "jumped" because absorbed in the "tails" being cut. Thanks for signalling that. – G Cab Jan 4 '17 at 18:52
• @Hurkyl: now should be better, thanks again for your comment – G Cab Jan 4 '17 at 19:30

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.

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

• Why does last piece of product (1-1/n^2)? I thought (1-1/(n+1)^2) – Yas Jan 4 '17 at 12:12
• @Yas, the answer has been edited to change that, but know that the two forms are equivalent. Your sum of $$\sum_{n=1}^\infty \log \left( 1 - \frac{1}{(n+1)^2} \right)$$ can be re-indexed by $m = n+1$ to get $$\sum_{m=2}^\infty \log \left( 1 - \frac{1}{m^2} \right)$$ – CodeLabMaster Jan 5 '17 at 5:41

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

• What's $DH$ on the equals sign? – tilper Jan 4 '17 at 19:35
• @tilper: De L'Hopital rule. – Jack D'Aurizio Jan 4 '17 at 19:56
• @JackD'Aurizio What does 'De' mean ? – A---B Jan 5 '17 at 7:18
• @A---B L'Hôpital's full name was/is Guillaume de L'Hospital/Guillaume de L'Hôpital. The d in de shouldn't be capitalized, so dH would be more appropriate. – Frank Vel Jan 5 '17 at 10:52

\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.

• Btw, I do not think that the edit by someone else was an improvement. – Carsten S Jan 6 '17 at 13:43