Alternative proof that $\frac{\sqrt[3]3}{\sqrt[2]2}\cdot\frac{\sqrt[5]5}{\sqrt[4]4}\cdot\frac{\sqrt[7]7}{\sqrt[6]6}\cdot ...=2^{-\gamma+\log\sqrt2}$ 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.
 A: 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.
