Question

Sum the series :
$2^{1/4}\cdot 4^{1/8}\cdot 8^{1/16}\cdot {16}^{1/32}....$ is equal

Answer 1

Consider the given series.

$2^{1/4}{.4}^{1/8}{.8}^{1/16}{.16}^{1/32}......$

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

$S=2^{(1/4+2/8+3/16+4/32+.......)}$

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

Let

$T=1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+.....$ (Infinity AGP series with a=1, r=1/2, d=1)

Therefore,

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

$T=\frac{1}{1-\frac{1}{2}}+\frac{1\times \frac{1}{2}}{{(1-\frac{1}{2})}^2}$

$T=\frac{1}{\frac{1}{2}}+\frac{\frac{1}{2}}{{\left(\frac{1}{2}\right)}^2}$

$T=2+2=4$

Therefore,

$S=2^{1/4\times 4}$

$S=2^1=2$

Hence, this is the answer.