Find the $\lim_{n \to \infty} \left(\sum_{r=1}^n \frac{1}{x_r}\right)$

Question

If $\{x_n\}$ is a sequence of real numbers with $$\ x_1=\frac{2014}{3}, \; x_{n+1}= x_n \left(1+\frac{1}{x_n-\lfloor x_n\rfloor}\right )\; , n \in \Bbb N$$ Find $$\lim_{n \to \infty}\left(\frac{1}{x_1}+\frac{1}{x_2}+ \cdots +\frac{1}{x_n}\right)$$ I think that we should write $x-\lfloor x \rfloor$ as a fractionary part but I can't handle it. Also the sequence is increasing. Please help me!

Answer 1

This looks impossible, but in fact $2014\equiv1\pmod3$ so $x_2=4x_1$ and since $4\equiv1\pmod3$ it will follow that each $$x_{n+1}=4x_n=\frac{4^{n+1}}{4^n}x_n$$ So $$\frac{x_{n+1}}{4^{n+1}}=\frac{x_n}{4^n}=\frac{x_1}4=\frac{1007}6$$ Then $$x_n=\frac{1007}64^n$$ So $$\sum_{n=1}^{\infty}\frac1{x_n}=\sum_{n=1}^{\infty}\frac6{1007}4^{-n}=\frac{\frac6{1007}4^{-1}}{1-4^{-1}}=\frac2{1007}$$