Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}=2^{\frac{1}{4}}.4^{\frac{2}{8}}.8^{\frac{3}{16}}.{16}^{\frac{4}{32}}...\infty \\ =2^{\left(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\frac{3}{16}\frac{4}{32}.\infty \right)} \\ =2^{\left(\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+...\infty \right)} \\ =2^{\frac{1}{2^2}\left\{1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}...\infty \right\}} \\ =2^{\frac{1}{2^2}\left\{\frac{1}{1-{\displaystyle \frac{1}{2}}}+\frac{1.{\displaystyle \frac{1}{2}}}{{\left(1-\frac{1}{2}\right)}^2}\right\}} \\ =2^{\frac{1}{2^2}\left\{2+2\right\}} \\ =2^1 \\ =2=\mathrm{RHS} \end{gathered}$$