I've been tasked with finding the upper and lower bounds of the element:
$A = sin(\frac{\pi.n}{2n+3}) | n\in\mathbb{N}$
I think I have found the upper bound by doing:
$\lim_{n\to +\infty} sin(\frac{\pi.n}{2n+3}) = \lim_{n\to +\infty} sin(\frac{\pi.n}{2n}) = \lim_{n\to +\infty} sin(\frac{\pi}{2})= 1$
And since every point of $sin(x)$ is confined within [-1,1], the upper bound can only be 1
But I'm completely stumped as to how I can find the lower bound.
I tried proving that the derivative does not equal to 0 for $n \in \mathbb{N}$, showing that it is only rising and thus, has 0 as a lower bound since A = 0 for n = 0, and it's where I'm stuck:
$\frac{3\pi cos(\frac{\pi n}{2n + 3})}{(2n+3)^2} = 0$
I can't escape having to deal with either $cos(\frac{\pi n}{2n + 3})$ or $sin(\frac{\pi n}{2n+3})$ as a single entity and I don't know how I can solve the derivative.
Best Answer
Notice that for all $n \in \mathbb{N}$ : $0\leq\frac{\pi n}{2 n+3}< \pi/2$.
Since that $\sin(x)$ is strictly increasing function and is continuous in the interval $[0,\pi/2]$,
you can automatically deduce that: $0=\sin(0)\leq\sin(\frac{\pi n}{2 n+3})< \sin(\pi/2)=1$