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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.

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    Older post about the same sum: [Prove that $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ converges and find its sum](http://math.stackexchange.com/q/1676164) and [Show that $\sum\limits_{k=1}^{\infty}\ln\left(\frac{k(k+2)}{(k+1)^2}\right) = -\ln(2)$?](http://math.stackexchange.com/q/1733923). Found [using Approach0](https://approach0.xyz/search/?q=%24%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20%5Cln%5Cleft(%5Cfrac%7Bn(n%2B2)%7D%7B(n%2B1)%5E2%7D%5Cright)%24&p=1). – 2017-01-05
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    I am amazed by the number of views and votes here. It doesn't make any sense. – 2017-01-05
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    @ZaidAlyafeai This usually happens when a question appears in [network-wide hot questions list](http://meta.math.stackexchange.com/tags/hot-questions-list/info). 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?](http://meta.math.stackexchange.com/q/20482) – 2017-01-05
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    @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 ? – 2017-01-05

6 Answers 6

30

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

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    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. – 2017-01-05
23

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

8

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

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    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. – 2017-01-04
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    @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. – 2017-01-04
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    @Hurkyl: now should be better, thanks again for your comment – 2017-01-04
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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.

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    Why does last piece of product (1-1/n^2)? I thought (1-1/(n+1)^2) – 2017-01-04
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    @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 $$\sum_{m=2}^\infty \log \left( 1 - \frac{1}{m^2} \right)$$ – 2017-01-05
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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: $$ \sum_{n\geq 1}\log\frac{n(n+2)}{(n+1)^2}=\log\prod_{n\geq 2}\left(1-\frac{1}{n^2}\right)=\log\lim_{x\to 1}\frac{\sin(\pi x)}{\pi x(1-x^2)}\stackrel{dH}{=}\color{red}{-\log 2}.\tag{2} $$

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    What's $DH$ on the equals sign? – 2017-01-04
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    @tilper: **D**e L'**H**opital rule. – 2017-01-04
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    @JackD'Aurizio What does 'De' mean ? – 2017-01-05
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    @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. – 2017-01-05
5

\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.

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    Btw, I do not think that the edit by someone else was an improvement. – 2017-01-06