Prove that the sequence is monotone and bounded above

**alexmahone** · January 15th 2012, 03:21 PM

Using calculus, prove (for $n\ge 2$ ) that $A_n=2^{n-1}\sin \frac{\pi}{2^{n-1}}$ is monotone (increasing) and bounded above. Find its limit.

**emakarov** · January 16th 2012, 04:10 AM

Expand sin(x) as x + O(x); then the limit of $A_n$ becomes obvious. To prove monotonicity and boundedness, expand $A_n$ as $a_0-a_1(n)+a_2(n)-\dots$ . This series satisfies the conditions to the alternating series test. From the info on that page, $A_n\le a_0$ . Also, $A_n\in[a_0-a_1(n),a_0-a_1(n)+a_2(n)]$ . You can show that $a_0-a_1(n+1)>a_0-a_1(n)+a_2(n)$ , so $A_{n+1}>A_n$ .

Another way to show monotonicity is to consider $f(x)=x\sin(1/x)$ and show that its derivative is eventually positive (also using series).

**chisigma** · January 16th 2012, 06:11 AM

Setting $x=2^{1-n}$ You have...

$\lim_{n \rightarrow \infty} 2^{n-1}\ \sin \frac{\pi}{2^{n-1}} = \lim_{x \rightarrow 0} \frac{\sin \pi x}{x} = \pi$

Kind regards

$\chi$ $\sigma$

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