Question
Prove that: $\Big(2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty\Big)=2$

Answer

$2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty$
$=2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty$
$=\Big(\frac14+\frac18+\frac{3}{16}+\frac{4}{32}+\ \dots\infty\Big)$
$=2$
$=2^{5}\cdots(1)$
$\text{S}=\frac{1}{4}+\frac28+\frac{3}{16}+\frac{4}{32}+\ \dots\infty$
$\text{S}=\Big(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\dots\infty\Big)2$
$\frac{\text{S}}{2}=\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\dots\infty$
$=\frac{\frac{1}{4}}{1-\frac12}$
$=\frac{1}{4}\times\frac21$
$\text{S}=\frac12$
$\text{S}=1$
Thus, $2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots=2^1=2$

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