Question

Prove that: (2^1/4 . 4^1/8 . 8^1/16. 16^1/32 ... ∞) = 2.

Answer 1

\[\text { LHS } = 2^\frac{1}{4} . 4^\frac{2}{8} . 8^\frac{3}{16} . {16}^\frac{4}{32} . . . \infty \]

\[ = 2^\left( \frac{1}{4} + \frac{2}{8} + \frac{3}{16}\frac{3}{16}\frac{4}{32} . \infty \right) \]

\[ = 2^\left( \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \frac{4}{2^5} + . . . \infty \right) \]

\[ = 2^\frac{1}{2^2}\left\{ 1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} . . . \infty \right\} \]

\[ = 2^\frac{1}{2^2}\left\{ \frac{1}{1 - \frac{1}{2}} + \frac{1 . \frac{1}{2}}{\left( 1 - \frac{1}{2} \right)^2} \right\} \]

\[ = 2^\frac{1}{2^2}\left\{ 2 + 2 \right\} \]

\[ = 2^1 \]

\[ = 2 = \text { RHS }\]