sqrt of sqrt
Posted: Fri Jul 09, 2021 2:34 am
Calculate $2\sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}$
\begin{equation}
\begin{split}
2\sqrt{2\sqrt[5]{2\sqrt[8]{2\sqrt[11]{2 \cdots}}}}&=2^{1+\frac12+\frac1{2\cdot5}+\frac1{2\cdot5\cdot8}+\frac1{2\cdot5\cdot8\cdot11}+\cdots}
\end{split}
\end{equation}
Here, let
\begin{equation}
\begin{split}
S&=1+\frac12+\frac1{2\cdot5}+\frac1{2\cdot5\cdot8}+\frac1{2\cdot5\cdot8\cdot11}+\cdots\\
&=1+\frac{\frac13}{\frac23}+\frac{(\frac13)^2}{(\frac23)\cdot(\frac23+1)}+\frac{(\frac13)^3}{(\frac23)\cdot(\frac23+1)\cdot(\frac23+2)}+\cdots\\
&=\sum_{k=0}^{\infty}\frac{\Gamma(\frac23)}{\Gamma(\frac23+k)}\left(\frac13\right)^k
\end{split}
\end{equation}
If we let \[f(x,n)=\sum_{k=0}^{\infty}\frac{\Gamma(n)}{\Gamma(n+k)}x^k\] Then by Wolframalpha,
\[f(x,n)=e^x(n-1)x^{1-n}\gamma(n-1,x)\]
Where $\gamma(s,x)$ is the Lower Incomplete Gamma function defined as
\[\gamma(s,x)=\int_0^xt^{s-1}e^{-t}dt\]
So, by substituting $n=\frac23,x=\frac13$, \[f\left(\frac23,\frac13\right)=-\frac13\sqrt[3]{\frac{e}3}\gamma\left(-\frac13,\frac13\right)\]
So, our answer is, \[\boxed{2^{-\frac13\sqrt[3]{\frac{e}3}\gamma(-\frac13,\frac13)}}\]