# Prove that the sequence is monotone and bounded above

• Jan 15th 2012, 03:21 PM
alexmahone
Prove that the sequence is monotone and bounded above
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.
• Jan 16th 2012, 04:10 AM
emakarov
Re: Prove that the sequence is monotone and bounded above
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).
• Jan 16th 2012, 06:11 AM
chisigma
Re: Prove that the sequence is monotone and bounded above
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

$\chi$ $\sigma$