$2^{1/4}\cdot 4^{1/8}\cdot 8^{1/16}\dots \mathrm{\infty }\text{ is equal to }$

$S=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\dots \mathrm{\infty } \to \left(1\right)$

$\begin{array}{l}\frac{1}{2}\mathrm{S}=\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\dots \mathrm{\infty }-\cdots \to \left(2\right)\\ \left(1\right)-\left(2\right)\Rightarrow \\ \mathrm{S}-\frac{1}{2}\mathrm{S}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+......\\ =\frac{\frac{1}{4}}{1-1/2}=1/2 \left(sum of infinite terms in G.P=\frac{a}{1-r}\right)\\ \frac{1}{2}\mathrm{S}=\frac{1}{2}\to \mathrm{S}=1\\ given sum=2^{\mathrm{S}}=2^1=2.\end{array}$