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Show that $$\prod_{n=2}^{+\infty }n^{(-1)^{n+1}/n}=\frac{\sqrt[3]{3}}{\sqrt[2]{2}}\cdot \frac{\sqrt[5]{5}}{\sqrt[4]{4}}\cdot \frac{\sqrt[7]{7}}{\sqrt[6]{6}}\cdot \frac{\sqrt[9]{9}}{\sqrt[8]{8}}\cdot \frac{\sqrt[11]{11}}{\sqrt[10]{10}}\cdot \frac{\sqrt[13]{13}}{\sqrt[12]{12}}\cdot ...=2^{-\gamma +\log \sqrt{2}}$$

The proof I know is what I will describe later, but will there be another way?

My proof: I need to prove that: $$\sum\limits_{n=2}^{+\infty }{\left( -1 \right)^{n+1}\frac{\log n}{n}}=\left( \log 2-2\gamma \right)\frac{\log 2}{2}$$ This is equivalent to proving $$\eta'(1)=\frac{\log(2)^2}{2}-\gamma \log(2)$$ where $$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\dotsc$$ is the Dirichlet Eta Function. It is well-known that $\eta(1)=\log(2)$ so we are interested in the linear coefficient in the power series of $\eta(s)$ around $s=1$. Our proof needs two ingredients:

  • The identity $\eta(s)=(1-2^{1-s}) \zeta(s)$ where $$\zeta(s)=1+\frac{1}{2^s}+\frac{1}{3^s}+\dotsc$$ is the Riemann Zeta Function.

  • The asymptotic expansion $\zeta(s)=\frac{1}{s-1}+\gamma+\mathcal{O}\left(s-1\right)$ where $s \to 1$.

We have: $$\begin{align*} \eta(s)&=(1-2^{1-s})\zeta(s)\\ &=\left(\log(2)(s-1)-\frac{\log(2)^2}{2} (s-1)^2+\mathcal{O}\left((s-1)^3\right)\right)\left(\frac{1}{s-1}+\gamma+\mathcal{O}\left(1-s\right)\right)\\ &=\log(2)+\left(\gamma \log(2)-\frac{\log(2)^2}{2}\right)(s-1)+\mathcal{O}\left((s-1)^2\right) \end{align*}$$ which is exactly what we wanted to prove.

Did
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whitexlotus
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  • @Oleg567 edited ^^ – whitexlotus Aug 20 '17 at 19:43
  • Frullani's integral? $$\ln(n)=\int_0^\infty\frac{e^{-x}-e^{-nx}}x~\mathrm dx$$ – Simply Beautiful Art Aug 20 '17 at 20:00
  • Apply Frullani's integral: $$\ln(n)=\int_0^\infty\frac{e^{-x}-e^{-nx}}x~\mathrm dx$$ Thus, \begin{align} \sum_{n=2}^\infty(-1)^{n+1}\frac{\ln(n)}n &=\int_0^\infty\sum_{n=2}^\infty(-1)^{n+1}\frac{e^{-x}-e^{-nx}}{nx}~\mathrm dx \&=\int_0^\infty\frac{e^{-x}\ln(2)-\ln(1+e^{-x})}x~\mathrm dx \&=\lim_{(r,R)\to(0^+,\infty)}\int_r^R\frac{\ln(2)}{xe^x}-\frac{\ln(2)}{e^x-1}+\frac{\ln(2)}{e^x-1}-\frac{\ln(1+e^x)}x+1~\mathrm dx \&=\lim_{(r,R)\to(0^+,\infty)}\ln(2)\left(\gamma+\frac12\ln\left(1+\frac1{e^x-1}\right)\right)+x\bigg]_r^R-\int_r^R\frac{\ln(1+e^x)}x~\mathrm dx \end{align} – Simply Beautiful Art Aug 20 '17 at 20:15
  • Hopefully someone can find this useful. I'm tired, and if noone picks this up by the time I'm back, I'll try to continue – Simply Beautiful Art Aug 20 '17 at 20:18
  • Here is some related stuff. – Simply Beautiful Art Aug 20 '17 at 20:19
  • Oh, I dug up an old posting which solves this problem in calculus level: https://math.stackexchange.com/questions/40998/the-sum-of-1n-frac-ln-nn – Sangchul Lee Aug 23 '17 at 14:04
  • https://math.stackexchange.com/questions/40998/the-sum-of-1n-frac-ln-nn/2405295#2405295 here other form – whitexlotus Aug 25 '17 at 03:01

1 Answers1

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Carrying on SBA's idea, Frullani's theorem leads to:

$$\begin{eqnarray*}\sum_{n\geq 1}(-1)^{n+1}\frac{\log n}{n}&=&\int_{0}^{+\infty}\sum_{n\geq 1}(-1)^{n+1}\frac{e^{-x}-e^{-nx}}{nx}\,dx\\&=&\int_{0}^{+\infty}\frac{\log(2)e^{-x}-\log(1+e^{-x})}{x}\,dx\\&\stackrel{IBP}{=}&-\log(2)\int_{0}^{+\infty}e^{-x}\log(x)\,dx+\int_{0}^{+\infty}\frac{\log(x)\,dx}{e^x+1}\\&=&-\log(2)\,\Gamma'(1)+\left.\frac{d}{ds}\int_{0}^{+\infty}\frac{x^s\,dx}{e^x+1}\right|_{s=0^+}\\&=&\gamma\log(2)+\left.\frac{d}{ds}\left((1-2^{-s})\Gamma(s+1)\zeta(s+1)\right)\right|_{s=0^+}\end{eqnarray*}$$ On the other hand, in a right neighbourhood of the origin we have: $$ 1-2^{-s} = \log(2) s-\frac{\log^2(2)}{2}s^2 +O(s^3) $$ $$ \Gamma(s+1) = 1-\gamma s+\frac{\pi^2+6\gamma^2}{12}s^2 +O(s^3) $$ $$ \zeta(s+1) = \frac{1}{s}+\gamma -\gamma_1 s +O(s^2) $$ hence: $$ (1-2^{-s})\Gamma(s+1)\zeta(s+1) = \log(2)-\frac{\log^2(2)}{2}s+O(s^2)$$ and: $$ \sum_{n\geq 1}(-1)^{n+1}\frac{\log n}{n}=\frac{1}{2}\log^2(2)-\gamma\log(2) $$ as wanted.

Jack D'Aurizio
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