# Two sequences defined by recurrence relations satisfy $x_n/y_n<\sqrt{7}$ for all $n$

Let $(x_n)_{n\geq 1}$ and $(y_n)_{n\geq 1}$ be two sequences such that: $$x_{n+1}=x_n^2+1 \quad \text{ and } \quad y_{n+1}=x_n y_n$$ with $x_1=2$ and $y_1=1$

Prove that for all $n$
$$\dfrac{x_n}{y_n} < \sqrt{7}.$$

Can I have any help with this situation?

• Initial values for $x_0$ and $y_0$? – Petite Etincelle Aug 19 '14 at 22:49
• no there is no initial values – Student Aug 19 '14 at 22:53
• At least you should assume $\frac{x_0}{y_0} <\sqrt{7}$ – Petite Etincelle Aug 19 '14 at 22:55
• even if we assume that you are thinking to solve it using recurrence but even u will face a problem by proving that – Student Aug 19 '14 at 22:58
• Gotta go somewhere, don't have time to think this through as an answer, but how about starting with$$\frac{x_{n+1}}{y_{n+1}}=\frac{x_n}{y_n}+\frac{1}{x_ny_n}\ ?$$ – David Aug 19 '14 at 23:00

Given that: $$\frac{x_{n+1}}{y_{n+1}}=\frac{x_n}{y_n}+\frac{1}{x_n y_n}$$ we have: $$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n y_n}+\frac{1}{x_{n-1} y_{n-1}}+\ldots+\frac{1}{x_1 y_1}+\frac{x_1}{y_1}$$ or just: $$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n\cdot\ldots\cdot x_1}+\frac{1}{x_{n-1}\cdot\ldots\cdot x_1}+\ldots+\frac{1}{x_1}+2.$$ Hence we need to prove that: $$\sum_{m=1}^{+\infty}\frac{1}{\prod_{n=1}^{m}x_n}\leq\sqrt{7}-2.\tag{1}$$ The form of the last inequality strongly suggest a continued fraction approach.
However, we can just notice that $x_1=2,x_2=5$ and for any $n\geq 3$ we have: $$x_n\geq 5^{2^{n-2}},$$ giving that the LHS of $(1)$ is less than $$\frac{1}{2}+\frac{1}{10}+\frac{1}{10}\sum_{k=1}^{+\infty}\frac{1}{5^{2^k}}\leq \frac{3}{5}+\frac{1}{10}\sum_{k=1}^{+\infty}\frac{1}{5^{k+1}}=\frac{121}{200}.$$ Since $\frac{521}{200}<\sqrt{7}$, we're done.