Question

Show that $(2^{1/4}\cdot 4^{1/8}\cdot 8^{1/6}\cdot {16}^{1/32}-\dots \infty )=2$

Answer 1

$S=2^{\frac{1}{4}}{2}^{\frac{2}{8}}{2}^{\frac{3}{16}}{2}^{\frac{4}{32}}....$

$S=2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}....}$

$S=2^{\frac{1}{4}(1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}....)}$

$S=2^{\frac{1}{4}\left(t\right)}$

$t=1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}....$ (Infinite AGP with $a=1,r=\frac{1}{2},d=1$)

$t=\frac{a}{(1-r)}+\frac{dr}{(1-r{)}^2}$

$t=2+2=4$

$S=2$