JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Sequences and Series

Rewrite each factor as a power of $2$: $2^{1/4}$, then $4^{1/16}=2^{2/16}=2^{1/8}$, then $8^{1/48}=2^{3/48}=2^{1/16}$, then $16^{1/128}=2^{4/128}=2^{1/32}$, and so on. The whole product is therefore $2$ raised to the sum of the exponents $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots$. That is a geometric series with first term $\frac{1}{4}$ and common ratio $\frac{1}{2}$, so its sum is $\frac{1/4}{1-1/2}=\frac{1}{2}$. The infinite product thus equals $2^{1/2}=\sqrt{2}$.