The problem I had was whether this series converges or diverges.
What I did was say that,
for every
and since diverges, so must
The problem here is that, I do not know how to prove that
although it seems plausible.
If you can prove that for , then you can take the arcsin of both sides and replace x with .
To prove that for , first note that it is obvious for . For x<1, use the mean value theorem for on [0,x]: for some c in the interval [0, x], so .
- Hollywood
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. does get proven in the 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 , by the way. Sorry about that.
- Hollywood
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
Once you've proved that the derivative of is , you're free to use it to prove other stuff.
You use to prove that the derivative of is , but the limit is derived from geometric arguments. The fact that is not involved in the process.
- Hollywood