Sum of the Sines of the Ratios of Fibonnaci Numbers

Question

Saw an online a proof that $$\lim_{n\to \infty}\sum_{k=1}^{n}\sin\left(\frac{k}{n^2}\right)=\frac{1}{2}$$ that utilized the Squeeze Theorem and the fact that $x-x^3\leq \sin(x)\leq x$ for small $x$. This got me thinking of the sum $$\lim_{n\to \infty}\sum_{k=1}^{n}\sin\left(\frac{F_k}{F_n}\right)$$ where $F_1=F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ for $n>2$. Testing partial sums on Wolfram Alpha confirms that such a series converges, and seems to approach around 2.4088209. I assume this might use similar tricks involving the golden ratio, but the same Squeeze Theorem trick doesn't seem to be useful here (it only really helped me show that the sum must be less that $\Phi^2$). Any one have any thoughts or methods on how to solve this?

Answer 1

Hint: write the sum "backwards", as $$\sum_k \ sin({F_{(n-k)} \over F_n})$$ and note that for large $n$, ${F_{(n-k)} \over F_n}\approx\phi^{-k}$