How to apply Stolz–Cesàro theorem in this question? $$\lim _{n \rightarrow \infty}\left(\frac{2}{2^{2}-1}\right)^{\frac{1}{2^{n-1}}} \left (\frac{2^{2}}{2^{3}-1}\right)^{\frac{1}{2^{n-2}}} \cdots\left(\frac{2^{n-1}}{2^{n}-1}\right)^{\frac{1}{2}}$$
I don't know how to continue after taking the logarithm of it, after that.
This is an exercise to find the limit using Stolz's method.
 A: Note
\begin{eqnarray}
&&\ln\bigg[\left(\frac{2}{2^{2}-1}\right)^{\frac{1}{2^{n-1}}} \left (\frac{2^{2}}{2^{3}-1}\right)^{\frac{1}{2^{n-2}}} \cdots\left(\frac{2^{n-1}}{2^{n}-1}\right)^{\frac{1}{2}}\bigg] \\
&=&\sum_{k=2}^n\frac{1}{2^{n-k+1}}\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)\\
&=&\frac{\sum_{k=2}^n2^k\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)}{2^{n+1}}.
\end{eqnarray}
By Stolz's theorem, one has
\begin{eqnarray}
&&\lim_{n\to\infty}\ln\bigg[\left(\frac{2}{2^{2}-1}\right)^{\frac{1}{2^{n-1}}} \left (\frac{2^{2}}{2^{3}-1}\right)^{\frac{1}{2^{n-2}}} \cdots\left(\frac{2^{n-1}}{2^{n}-1}\right)^{\frac{1}{2}}\bigg] \\
&=&\lim_{n\to\infty}\frac{\sum_{k=2}^n2^k\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)}{2^{n+1}}\\
&=&\lim_{n\to\infty}\frac{\sum_{k=2}^{n+1}2^k\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)-\sum_{k=2}^n2^k\ln\bigg(\frac{2^{k-1}}{2^{k}-1}\bigg)}{2^{n+2}-2^{n+1}}\\
&=&\lim_{n\to\infty}\frac{2^{n+1}\ln\bigg(\frac{2^{n}}{2^{n+1}-1}\bigg)}{2^{n+1}}\\
&=&\ln(\frac12)
\end{eqnarray}
and hence
$$\lim _{n \rightarrow \infty}\left(\frac{2}{2^{2}-1}\right)^{\frac{1}{2^{n-1}}} \left (\frac{2^{2}}{2^{3}-1}\right)^{\frac{1}{2^{n-2}}} \cdots\left(\frac{2^{n-1}}{2^{n}-1}\right)^{\frac{1}{2}}=\frac12.$$
