Using calculus, prove (for $\displaystyle n\ge 2$) that $\displaystyle A_n=2^{n-1}\sin \frac{\pi}{2^{n-1}}$ is monotone (increasing) and bounded above. Find its limit.
Using calculus, prove (for $\displaystyle n\ge 2$) that $\displaystyle A_n=2^{n-1}\sin \frac{\pi}{2^{n-1}}$ is monotone (increasing) and bounded above. Find its limit.
Expand sin(x) as x + O(x); then the limit of $\displaystyle A_n$ becomes obvious. To prove monotonicity and boundedness, expand $\displaystyle A_n$ as $\displaystyle a_0-a_1(n)+a_2(n)-\dots$. This series satisfies the conditions to the alternating series test. From the info on that page, $\displaystyle A_n\le a_0$. Also, $\displaystyle A_n\in[a_0-a_1(n),a_0-a_1(n)+a_2(n)]$. You can show that $\displaystyle a_0-a_1(n+1)>a_0-a_1(n)+a_2(n)$, so $\displaystyle A_{n+1}>A_n$.
Another way to show monotonicity is to consider $\displaystyle f(x)=x\sin(1/x)$ and show that its derivative is eventually positive (also using series).
Setting $\displaystyle x=2^{1-n}$ You have...
$\displaystyle \lim_{n \rightarrow \infty} 2^{n-1}\ \sin \frac{\pi}{2^{n-1}} = \lim_{x \rightarrow 0} \frac{\sin \pi x}{x} = \pi$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$