I want to find the growth rate of $f(n)$ as $\lim \limits_{n\to \infty} f(n)$
I can't find an other function that bounds this one in some form. Also the algebra manipulations involved here have confused me. I would appreciate a little help. Thanks.
2026-04-24 07:05:12.1777014312
Asymptotic growth of $f(n) = \left( \sum\limits_{i=1}^n \frac{1}{4^i} \right)^n $?
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Unless I am missing something, $$(f(n))^{1/n} = \frac{1-4^{-n-1}}{1-1/4} - 1 = \frac{4-4^{-n}}{3} - 1 = \frac{1-4^{-n}}{3}. $$ Therefore, $$f(n) = \frac{(1-4^{-n})^n}{3^n},$$ the numerator converges to $1$ and the denominator converges to $+\infty$, hence $\lim_{n\to\infty} f(n)=0$.