Tìm $\lim{u_n}$

Cho dãy số $(u_n)$ biết :
\[u_n=\frac{2}{\sqrt2}.\frac{2}{\sqrt{2+\sqrt{2}}}.....\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+....\sqrt{2}}}}}\]
Tìm $\lim{u_n}$
Lời giải
Trước hết ta chứng minh công thức
$$\sqrt{2+\sqrt{2+...+\sqrt{2}}} \text{ (n-căn)}=2\cos \dfrac{\pi}{2^{n+1}}.$$
$u_n$ sau khi rút gọn sẽ trở thành
$$u_n=\dfrac{1}{\cos \dfrac{\pi}{4} \cos \dfrac{\pi}{8}...\cos \dfrac{\pi}{2^{n+1}}}$$
Xét $v_n=\dfrac{1}{u_n}$. Ta có:
$$\begin{align*}
v_n\sin \dfrac{\pi}{2^{n+1}}&=\cos \dfrac{\pi}{4} \cos \dfrac{\pi}{8}...\cos \dfrac{\pi}{2^{n+1}}\sin \dfrac{\pi}{2^{n+1}}\\
&=\dfrac{1}{2}\cos \dfrac{\pi}{4} \cos \dfrac{\pi}{8}...\cos \dfrac{\pi}{2^{n}}\sin \dfrac{\pi}{2^n}\\
&= ....\\
&= \frac{1}{2^n}sin \frac{\pi}{2}=\frac{1}{2^n}.
\end{align*}$$
suy ra $v_n=\dfrac{1}{2^n \sin \dfrac{\pi}{2^{n+1}}}\Rightarrow u_n=\dfrac{1}{v_n}=2^n \sin \dfrac{\pi}{2^{n+1}}.$
Chuyển qua tính giới hạn, ta thu được
$$\lim_{n \to \infty}u_n=\lim_{n \to \infty}2^n \sin \dfrac{\pi}{2^{n+1}}=\lim_{n \to \infty} \frac{\pi}{2}\dfrac{\sin \dfrac{\pi}{2^{n+1}}}{\dfrac{\pi}{2^{n+1}}}=\dfrac{\pi}{2}.$$