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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

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

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

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

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

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.

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

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