# Math Help - Series/limit problem

1. ## 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.

2. ## 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?

3. ## 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

4. ## 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.

5. ## 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 $arcsinx\geq x$

6. ## Re: Series/limit problem

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

7. ## 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

8. ## 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

9. ## 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.

10. ## 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

11. ## Re: Series/limit problem

The geometric argument proves that $sinx\leq{x}\leq{tanx}$. That is the purpose of the geometric argument.

12. ## 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