I ran across an infinite series that is allegedly from a Chinese math contest.



I thought perhaps this telescoped in some fashion. So, I wrote out


Separated the positive and negative using log properties:

$\ln(1/2)+\ln(11/12)+\ln(29/30)+......=\ln(\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot)$

$-(\ln(5/6)+\ln(19/20)+\ln(41/42)+......=-\ln(\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot) $

$\ln(\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot)-\ln(\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot)$

$=\displaystyle \ln\left(\frac{\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot}{\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot}\right)$

Maybe come up with a general term at the end of the partial sum? The terms in the numerator are $n=2,4,6,....$ and those in the denominator are $n=3,5,7,.....$

$\frac{N(N-1)-1}{N(N-1)}$. But, I always end up with a limit of 1. This then gives $\ln(1)=0$.

The series does converge. I managed to do some cancellations, but failed to wrap it up.

I thought maybe I was onto something. I suppose I am and not seeing it. What would be a good plan of attack for this one? Since it was in a contest, I assume it can be done. Any thoughts?

Thanks very much.


Put everything into an infinite product: $$\log\prod_{n=1}^\infty \frac{1-\frac{1}{2n(2n-1)}}{1-\frac{1}{(2n+1)2n}}=\log\prod_{n=1}^\infty\frac{8n^3-4n-1}{8n^3-4n+1}$$ Factor the polynomials and look at partial products: $$=\log\prod_{n=1}^{m-1}\frac{n+1/2}{n-1/2}\frac{n-\varphi/2}{n+\varphi/2}\frac{n-\Phi/2}{n+\Phi/2}$$ (NB $\varphi=\frac{1+\sqrt{5}}{2},\Phi=\frac{1-\sqrt{5}}{2}$.) Substitute using $\Gamma$'s properties: $$\log\left((2m+1)\frac{\Gamma(m-\varphi/2)\Gamma(-\varphi/2)^{-1}}{\Gamma(m+\varphi/2)\Gamma(\varphi/2)^{-1}}\frac{\Gamma(m-\Phi/2)\Gamma(-\Phi/2)^{-1}}{\Gamma(m+\Phi/2)\Gamma(\Phi/2)^{-1}}\right)$$ Evaluate the limit as $m\to\infty$ using Stirling's Formula: $$\log\left(\frac{2\Gamma(\varphi/2)\Gamma(\Phi/2)}{\Gamma(-\varphi/2)\Gamma(-\Phi/2)}\right).$$


Use $ \log\left(1 - \frac{1}{n(n-1)}\right) = \int_0^1 \frac{\mathrm{d} t}{n(1-n)+t}$.


$$ \begin{eqnarray} \sum_{n=2}^\infty \frac{(-1)^n}{n - n^2 + t} &=& \sum_{n=2}^\infty \frac{2 (-1)^n }{\sqrt{4 t+1}} \left(\frac{1}{2 n+\sqrt{4 t+1}-1}-\frac{1}{2 n-\sqrt{4 t+1}-1}\right) \\&=& \frac{1}{2 \sqrt{4 t+1}} \left( \psi ^{(0)}\left(-\frac{1}{4} \sqrt{4 t+1}-\frac{1}{4}\right)-\psi ^{(0)}\left(\frac{1}{4}-\frac{1}{4} \sqrt{4 t+1}\right) \right) \\ &+& \frac{1}{2 \sqrt{4 t+1}} \left(\psi ^{(0)}\left(\frac{1}{4} \sqrt{4 t+1}+\frac{1}{4}\right) -\psi ^{(0)}\left(\frac{1}{4} \sqrt{4 t+1}-\frac{1}{4}\right)\right) \end{eqnarray} $$ The latter comes about from $\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+a}\right) = \gamma + \psi^{(0)}(a+1)$, and the summation above was split into summation over even and odd integers.

Integrating this expression out produces: $$ \text{log$\Gamma $}\left(\frac{1}{2}-\frac{\phi}{2} \right)-\text{log$\Gamma $}\left(-\frac{1}{2}+\frac{\phi }{2}\right)-\text{log$\Gamma $}\left(-\frac{\phi }{2}\right)+\text{log$\Gamma $}\left(\frac{\phi }{2}\right)+\log (2) $$ where $\phi$ is Golden ratio.

Integration is trivial as $\frac{\mathrm{d} t}{\sqrt{4 t+1}} = \mathrm{d}\left(\frac{\sqrt{4 t+1}}{2}\right)$, and $\int \psi^{(0)}(u) \mathrm{d} u = \log\Gamma(u) + C$.

Numerical check in Mathematica:

In[85]:= N[
 Log[2] - LogGamma[-GoldenRatio/2] + 
  LogGamma[-(GoldenRatio/2) + 1/2] - LogGamma[GoldenRatio/2 - 1/2] + 
  LogGamma[GoldenRatio/2], 20]

Out[85]= -0.56655310975860303045 + 0.*10^-21 I

In[84]:= NSum[(-1)^n Log[1 - 1/(n (n - 1))], {n, 2, \[Infinity]}, 
 WorkingPrecision -> 20]

Out[84]= -0.566553109758603
  • $\begingroup$ 'Bravura' is the word that comes to mind! $\endgroup$ – TonyK Oct 1 '11 at 21:46

Combining every two terms and substituting $n\mapsto2n$, we get $$ \begin{align} \sum_{n=2}^\infty(-1)^n\ln\left(1-\frac{1}{n(n-1)}\right) &=\ln\left(\prod_{n=1}^\infty\frac{(2n+1)2n}{2n(2n-1)}\frac{(2n-\phi)(2n+\phi-1)}{(2n-\phi+1)(2n+\phi)}\right)\\ &=\ln\left(\prod_{n=1}^\infty\frac{n+1/2}{n-1/2}\frac{(n-\phi/2)(n+\phi/2-1/2)}{(n-\phi/2+1/2)(n+\phi/2)}\right) \end{align} $$ The partial product is $$ \begin{align} &\prod_{n=1}^N\frac{n+1/2}{n-1/2}\frac{(n-\phi/2)(n+\phi/2-1/2)}{(n-\phi/2+1/2)(n+\phi/2)}\\ &=(2N+1)\frac{\Gamma(N-\phi/2+1)}{\Gamma(1-\phi/2)}\frac{\Gamma(N+\phi/2+1/2)}{\Gamma(1/2+\phi/2)}\frac{\Gamma(3/2-\phi/2)}{\Gamma(N-\phi/2+3/2)}\frac{\Gamma(1+\phi/2)}{\Gamma(N+\phi/2+1)} \end{align} $$ Now we use the fact that $\displaystyle\lim_{x\to\infty}\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}=1$ to get that the limit of the partial product is $$ 2\frac{\Gamma(3/2-\phi/2)\Gamma(1+\phi/2)}{\Gamma(1-\phi/2)\Gamma(1/2+\phi/2)}=2\frac{\Gamma(1/2-\phi/2)\Gamma(\phi/2)}{\Gamma(-\phi/2)\Gamma(\phi/2-1/2)} $$ So the answer is $$ \ln\left(2\frac{\Gamma(1/2-\phi/2)\Gamma(\phi/2)}{\Gamma(-\phi/2)\Gamma(\phi/2-1/2)}\right) $$ Appendix: Searching on the web, the the closest to a reference for $$ \lim_{x\to\infty}\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}=1 $$ I could find was Gautschi’s Inequality in the Digital Library of Mathematical Functions and no proof of this inequality is given. Since I find it quite useful and it is a simple consequence of the log-convexity of $\Gamma$, I provide a proof here.

$\Gamma$ is the log-convex function for which $\Gamma(1)=1$ and $\Gamma(x+1)=x\Gamma(x)$. Thus, for $0\le\alpha\le1$, $$ \begin{align} \Gamma(x+\alpha) &\le\Gamma(x)^{1-\alpha}\Gamma(x+1)^\alpha\\ &=\Gamma(x)x^\alpha\tag{1} \end{align} $$ and $$ \begin{align} \Gamma(x)x^\alpha &=\Gamma(x+1)x^{\alpha-1}\\ &\le\Gamma(x+\alpha)^\alpha\Gamma(x+\alpha+1)^{1-\alpha}x^{\alpha-1}\\ &=\Gamma(x+\alpha)(x+\alpha)^{1-\alpha}x^{\alpha-1}\\ &=\Gamma(x+\alpha)(1+\alpha/x)^{1-\alpha}\tag{2} \end{align} $$ Combining $(1)$ and $(2)$, we get $$ (1+\alpha/x)^{\alpha-1}\le\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}\le1\tag{3} $$ By the Sandwich Theorem, $(3)$ yields $$ \lim_{x\to\infty}\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}=1\tag{4} $$ for $0\le\alpha\le1$. However, we can extend $(4)$ to any $\alpha$ using $$ \begin{align} \frac{\Gamma(x+\alpha+k)}{x^k\Gamma(x+\alpha)} &=\frac{(x+\alpha)(x+\alpha+1)(x+\alpha+2)\dots(x+\alpha+k-1)}{x^k}\\ &\to1\text{ as }x\to\infty\tag{5} \end{align} $$ or $$ \begin{align} \frac{\Gamma(x+\alpha-k)}{x^{-k}\Gamma(x+\alpha)} &=\frac{x^k}{(x+\alpha-k)(x+\alpha-k+1)(x+\alpha-k+2)\dots(x+\alpha-1)}\\ &\to1\text{ as }x\to\infty\tag{6} \end{align} $$

  • $\begingroup$ I see that my answer is similar to anon's, except that I use $\displaystyle\lim_{x\to\infty}\frac{\Gamma(x+\alpha)}{x^\alpha\Gamma(x)}=1$ instead of Stirling. $\endgroup$ – robjohn Oct 2 '11 at 1:28
  • $\begingroup$ Wow, thank you all very much. I was thinking along the lines of the latter two, but failed to see the wonderful solutions involving Gamma. I am familiar with the Digamma series as well, but did not think of that. I do not want to vote for anyone in particular because they are all great solutions. Do I have to?. $\endgroup$ – Cody Oct 2 '11 at 13:07
  • $\begingroup$ @rob: if you mouse over the "i" icons, you'll find a link to the bibliography; unfortunately, there seems to be no digital version of Gautschi's article anywhere... :( $\endgroup$ – J. M. is a poor mathematician Oct 2 '11 at 13:38
  • $\begingroup$ @J. M.: Thanks; I didn't notice the "i" at the right. You seem to be right about Gautschi's article and Laforgia's article simply references 6.1.46 in Abramowitz & Stegun. $\endgroup$ – robjohn Oct 2 '11 at 14:13
  • 1
    $\begingroup$ This is related, and so you might find the answers there interesting: A gamma function inequality. $\endgroup$ – Mike Spivey Oct 2 '11 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.