A piece of $\pi$

Calculus Level 5

$\lim_{n \to \infty} 2^{n+\frac{5}{2}} \sqrt{3-\sqrt{3+3T_{n}}}$

Define a sequence $T$ with $n^\text{th}$ term $T_{n}$ such that $T_{0} = 0$ and $T_{n} = \sqrt{2+T_{n-1}}$ . Then evaluate the limit above.

6\pi

2\pi

\frac{\pi}{2}

\pi^{2}

\pi

1 solution

Arturo Presa
Feb 12, 2016

Since $2\cos\frac{\pi}{2}=0$ and $2\cos \frac{\pi}{2^{n+1}}=\sqrt{2+2\cos \frac{\pi}{2^{n}}},$ then $T_n= 2\cos \frac{\pi}{2^{n+1}}.$ Denoting the limit of the question by $L,$ we have that $\begin{aligned} L&=\lim_{n \to \infty} 2^{n+\frac{5}{2}} \sqrt{3-\sqrt{3+6\cos \frac{\pi}{2^{n+1}}}}\\ &=\lim_{n \to \infty} 2^{n+\frac{5}{2}} \sqrt{3-\sqrt{3+6\cos \frac{\pi}{2^{n+1}}}}\cdot\frac{\sqrt{3+\sqrt{3+6\cos \frac{\pi}{2^{n+1}}}}}{\sqrt{3+\sqrt{3+6\cos \frac{\pi}{2^{n+1}}}}}\\ &=\lim_{n \to \infty} 2^{n+\frac{5}{2}} \sqrt{6-6\cos \frac{\pi}{2^{n+1}}}\cdot\lim_{n \to \infty} \frac{1}{\sqrt{3+\sqrt{3+6\cos \frac{\pi}{2^{n+1}}}}}\\ &= \lim_{n \to \infty} 2^{n+\frac{5}{2}} \sqrt{6-6\cos \frac{\pi}{2^{n+1}}} \cdot \frac{1}{\sqrt{6}}\\ &=\lim_{n \to \infty} 2^{\frac{3}{2}}\pi \sqrt{\frac{1-\cos \frac{\pi}{2^{n+1}}}{(\frac{\pi}{2^{n+1}})^2}}\\ &=\boxed{ 2\pi }\end{aligned}$

A piece of π \pi π

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A piece of $\pi$