# Find the smallest positive number k

For any positive number $$n$$, let $$a_n = \sqrt{2+\sqrt{2+{...+\sqrt{2+\sqrt 2}}}}$$ ($$2$$ appear $$n$$) and let $$k$$ is positive number such that $$\displaystyle\frac{1}{k}\leq\frac{3-a_{n+1}}{7-a_n}$$ for any positive number $$n$$, then find the smallest positive number $$k$$.

I have $$a_1=\sqrt 2$$ and $$a_{n+1}=\sqrt{2+a_n}, \forall n \in\mathbb N$$

Consider

$$a_1=\sqrt 2 \lt 2$$

$$a_2=\sqrt{2+a_1}\lt\sqrt{2+2}=2$$

$$a_3=\sqrt{2+a_2}\lt\sqrt{2+2}=2$$

Use Mathematical Induction, I conclude $$\sqrt 2\leq a_n\leq 2,\forall n\in\mathbb N$$

Thus, $$3-a_{n+1}\gt1$$ and $$7-a_{n}\gt 5$$

Since $$k\in\mathbb N$$, I have $$\displaystyle k\geq\frac{7-a_n}{3-a_{n+1}}=\frac{7-a_n}{3-\sqrt{2+a_n}}=3+\sqrt{2+a_n}=3+a_{n+1}$$

Hence, $$3+\sqrt 2\leq 3+a_{n+1}\leq 3+2=5$$

Therefore $$k=5$$

Please check my solution, Is it correct?, Thank you

• $a_n=2\cos\frac{\pi}{2^{n+1}}$. – Riemann Dec 18 '19 at 8:06

You have given a lower & upper bound for $$a_n$$. With this, you've determined a possible value for $$k$$, but you haven't shown it's necessarily the smallest such value of $$k$$.

Instead, note that $$a_{n+1} = \sqrt{2 + a_n}$$. Thus, you have

\begin{aligned} \frac{3-a_{n+1}}{7-a_n} & = \frac{3-\sqrt{2 + a_n}}{7-a_n} \\ & = \frac{(3-\sqrt{2 + a_n})(3 + \sqrt{2 + a_n})}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{9-(2 + a_n)}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{7 - a_n}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{1}{3 + \sqrt{2 + a_n}} \end{aligned}\tag{1}\label{eq1A}

Thus, if $$L$$ is the supremum of $$a_n$$, then $$k = 3 + \sqrt{2 + L}$$. To determine $$L$$, you can prove that $$a_n$$ is a strictly increasing sequence (I'll leave it to you fill in the details, such as what is shown in Solution verification: Prove by induction that $$a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n}$$ is increasing and bounded by $$2$$), with an upper bound as you've shown, so it must converge to limiting value of its supremum. To determine this value, use

\begin{aligned} L & = \sqrt{2 + L} \\ L^2 & = 2 + L \\ L^2 - L - 2 & = 0 \\ (L - 2)(L + 1) & = 0 \end{aligned}\tag{2}\label{eq2A}

Thus, since $$L \gt 0$$, you have $$L = 2$$, as you surmised. You also have

$$k = 3 + \sqrt{2 + 2} = 5 \tag{3}\label{eq3A}$$

which matches what you got.

• $a_n=2\cos\frac{\pi}{2^{n+1}}$. – Riemann Dec 18 '19 at 8:07