# Series/limit problem

• May 8th 2013, 06:12 AM
HK47
Series/limit problem
The problem I had was whether this series converges or diverges.

$\sum\limits_{n=10}^n arcsin\frac{\pi}{\sqrt{n}}$

What I did was say that,

$arcsin\frac{\pi}{\sqrt{n}}\geq\frac{1}{\sqrt{n}}$ for every $n$

and since $\sum\limits_{n=10}^n\frac{1}{\sqrt{n}}$ diverges, so must $\sum\limits_{n=10}^narcsin\frac{\pi}{\sqrt{n}}$

The problem here is that, I do not know how to prove that

$arcsin\frac{\pi}{\sqrt{n}}\geq\frac{1}{\sqrt{n}}$

although it seems plausible.
• May 8th 2013, 07:16 AM
HK47
Re: Series/limit problem
I did figure out how to solve this with the comparison test from reading older posts. But can it be done the way I described?
• May 8th 2013, 08:20 AM
hollywood
Re: Series/limit problem
If you can prove that $\sin{x}\le{x}$ for $x\ge{0}$, then you can take the arcsin of both sides and replace x with $\frac{1}{\sqrt{x}}$.

To prove that $\sin{x}\le{x}$ for $x\ge{0}$, first note that it is obvious for $x\ge{1}$. For x<1, use the mean value theorem for $\sin{x}$ on [0,x]: $\frac{\sin{x} - \sin{0}}{x - 0} = \cos{c}$ for some c in the interval [0, x], so $\frac{\sin{x}}{x}\ge\cos{c}\ge{1}$.

- Hollywood
• May 9th 2013, 11:54 PM
HK47
Re: Series/limit problem
I'm not sure about your use of the mean value theorem. Doesn't it just say, that there is a point, in that range, where the derivative is equal to the mean change of the functions value, not that the derivative is less than the mean change of value in the whole range, which would need to be proven. $sinx\leq x$ does get proven in the $lim_{x\to 0}\frac{sinx}{x}$however.
• May 10th 2013, 12:01 AM
HK47
Re: Series/limit problem
Oh, i just thought of something neat! Since arcsin(x) is the inverse function of sinx, then if the $sinx\leq x$ then the inverse function reflects sinx around $y=x$ , which would make $arcsinx\geq x$
• May 11th 2013, 12:39 AM
hollywood
Re: Series/limit problem
Quote:

Originally Posted by HK47
I'm not sure about your use of the mean value theorem. Doesn't it just say, that there is a point, in that range, where the derivative is equal to the mean change of the functions value, not that the derivative is less than the mean change of value in the whole range, which would need to be proven. $sinx\leq x$ does get proven in the $lim_{x\to 0}\frac{sinx}{x}$however.

That's exactly correct. At some point c between 0 and x, the derivative is equal to the mean change. But I'm using that to prove something about the mean change, not about the derivative. Since the derivative is less than or equal to 1, the mean change is less than or equal to 1. And since x was an arbitrary number between 0 and 1, the mean change is less than or equal to 1 for all x between 0 and 1.

I should have put $\frac{\sin{x}}{x} = \cos {c} \le 1$, by the way. Sorry about that.

- Hollywood
• May 14th 2013, 03:01 PM
HK47
Re: Series/limit problem
Thanks for the last post. It really made me think .. alot. Anyway, I think the reasoning is correct, but it is not a proof, you use the derivative of sinx, which is proven based on the same sinx is smaller than x, that you set out to prove :)
• May 15th 2013, 08:00 AM
hollywood
Re: Series/limit problem
No, the original problem was to compute the sum $\sum_{n=10}^\infty \sin^{-1}\frac{\pi}{\sqrt{n}}$, and $\frac{d}{dx}\sin{x}=\cos{x}$ should be proven by the time you're working on infinite series.

- Hollywood
• May 16th 2013, 03:00 PM
HK47
Re: Series/limit problem
Sorry, my last post was unclear. I am saying that in order to prove that $(sinx)'=cosx$ you must first prove $limit_{x\to\0}\frac{sin(x)}{x}$. And to prove $limit_{x\to\0}\frac{sin(x)}{x}$ you must first prove. $x\leq{sin(x)}$. So using $(sinx)'=cosx$ to prove $x\leq{sin(x)}$ doesnt function as a proof, as the proof is implicit in the derivative.
• May 16th 2013, 10:38 PM
hollywood
Re: Series/limit problem
Once you've proved that the derivative of $\sin{x}$ is $\cos{x}$, you're free to use it to prove other stuff.

You use $\lim_{x\to{0}}\frac{\sin{x}}{x}=1$ to prove that the derivative of $\sin{x}$ is $\cos{x}$, but the limit is derived from geometric arguments. The fact that $x\le\sin{x}$ is not involved in the process.

- Hollywood
• May 17th 2013, 12:32 AM
HK47
Re: Series/limit problem
The geometric argument proves that $sinx\leq{x}\leq{tanx}$. That is the purpose of the geometric argument.
• May 17th 2013, 07:45 AM
hollywood
Re: Series/limit problem
Ok, you're right. So we've actually already proven $\sin{x}\le{x}$ (let me get the inequality in the right direction this time) for $x\ge{0}$.

- Hollywood