Question

The value of $\underset{n\to \infty }{lim}{\left(\frac{1}{5}\right)}^{{\log }_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots \text{to n terms})}$ is​​​​​​​

Answer 1

Given,
$\underset{n\to \infty }{lim}{\left(\frac{1}{5}\right)}^{{\log }_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots \text{to n terms})}$
$={\left(\frac{1}{5}\right)}^{{\log }_{\sqrt{5}}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\text{to}\infty \text{terms})}$
$=5^{-2{\log }_5\left[\frac{1/4}{1-1/2}\right]}$
$=5^{{\log }_5{\left[\frac{1/4}{1-1/2}\right]}^{-2}}$
$={(\frac{1}{4}\times \frac{2}{1})}^{-2}={\left(2^{-1}\right)}^{-2}=2^2=4$