# challenging alternating infinite series involving $\ln$

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

Evaluate:

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

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

$\ln(1/2)-\ln(5/6)+\ln(11/12)-\ln(19/20)+\ln(29/30)-..............$

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

Then

$$\begin{eqnarray} \sum_{n=2}^\infty \frac{(-1)^n}{n - n^2 + t} &=& \sum_{n=2}^\infty \frac{2 (-1)^n }{\sqrt{4 t+1}} \left(\frac{1}{2 n+\sqrt{4 t+1}-1}-\frac{1}{2 n-\sqrt{4 t+1}-1}\right) \\&=& \frac{1}{2 \sqrt{4 t+1}} \left( \psi ^{(0)}\left(-\frac{1}{4} \sqrt{4 t+1}-\frac{1}{4}\right)-\psi ^{(0)}\left(\frac{1}{4}-\frac{1}{4} \sqrt{4 t+1}\right) \right) \\ &+& \frac{1}{2 \sqrt{4 t+1}} \left(\psi ^{(0)}\left(\frac{1}{4} \sqrt{4 t+1}+\frac{1}{4}\right) -\psi ^{(0)}\left(\frac{1}{4} \sqrt{4 t+1}-\frac{1}{4}\right)\right) \end{eqnarray}$$ 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:= N[
Log - LogGamma[-GoldenRatio/2] +
LogGamma[-(GoldenRatio/2) + 1/2] - LogGamma[GoldenRatio/2 - 1/2] +
LogGamma[GoldenRatio/2], 20]

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

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

Out= -0.566553109758603

• 'Bravura' is the word that comes to mind! – 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}

• 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. – robjohn Oct 2 '11 at 1:28
• 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?. – Cody Oct 2 '11 at 13:07
• @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... :( – J. M. is a poor mathematician Oct 2 '11 at 13:38
• @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. – robjohn Oct 2 '11 at 14:13
• This is related, and so you might find the answers there interesting: A gamma function inequality. – Mike Spivey Oct 2 '11 at 17:25