How do i find the convergence of $f(n)/2^{n}$

Question

I need to find the convergence of $\frac{f(n)}{2^{n}}$ where $f(n)$ is the $n^{th}$ element of the Fibonacci sequence. I have a very large N (>1000) and computing the value is numerically difficult. Any help is greatly appreciated! Thanks!

Answer 1

From here you know that $\sum_{n \ge 1} \frac{f(n)}{2^n} = 2$, therefore, we neccessary have $\lim_{n \to \infty} \frac{f(n)}{2^n} = 0$.

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My not yet finished attempt to show the limit statement goes like this: \begin{align} \frac{f(n)}{2^n} & = \frac{1}{2^n} \cdot \frac{\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n}{\sqrt{5}} = \frac{4^{-n}}{\sqrt{5}} \left( (1 + \sqrt{5})^n - (1 - \sqrt{5})^n \right) \\ & = \frac{4^{-n}}{\sqrt{5}} \sum_{k = 0}^{n} \binom{n}{k} (\sqrt{5})^k(1 - (-1)^n) \end{align} Now if we assume that $n$ is even we can continue: $$ \frac{f(n)}{2^n} = \frac{4^{-n}}{\sqrt{5}} \sum_{k = 0}^{\frac{n}{2}}2 \binom{n}{2k} (\sqrt{5})^{2k} = \frac{4^{-n}}{\sqrt{5}} \sum_{k = 0}^{\frac{n}{2}}2 \binom{n}{2k} (\sqrt{5})^{2k} = \frac{2^{1-2n}}{\sqrt{5}} \sum_{k = 0}^{\frac{n}{2}} \binom{n}{2k} 5^k $$ We'd now need some sort of asymptotic estimate to take care of the sum.