Prove the the power series converge uniformly I want to prove that $$sin(x) := \sum_{n = 0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$$ converges uniformly on any bounded interval $I$. 
I do not understand the concept of uniform convergence of series well. I am trying to apply a couple of Theorems (Cauchy Criterion, Weierstrass M-Test, etc.) but it leads me nowhere.
 A: Choose any interval, say $[x_1,x_2]$.  Then, note that in this interval, $|x^{2n+1}|\le \left(\max(|x_1|,|x_2|)\right)^{2n+1}$.  Therefore, we have
$$\left|\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{2n+1}}{(2n+1)!}\right|\le \sum_{n=1}^\infty \frac{\left(\max(|x_1|,|x_2|)\right)^{2n+1}}{(2n+1)!} \tag 1$$
The right-hand side of $(1)$ converges by the ratio test.  Therefore, by the Weierstrass M-Test, the series for $\sin x$ converges uniformly on $[x_1,x_2]$ for any (finite) values of $x_1$ and $x_2$.
A: We want to show that, for any $\epsilon$, there is a $\delta=\delta(\epsilon)$ for which $|\sin(x)-\sin(x+\delta)|<\epsilon$, for all $x\in I$.  So $\delta$ can't depend on $x$, but it can depend on $I$.
Let the interval $I=[-M,M]$ be all numbers between $-M$ and $M$, for some (large) $M$.
Look at $\frac1{(2n+1)!}|x^{2n+1}-(x+\delta)^{2n+1}|$  By the Mean Value Theorem, this equals $\frac{(2n+1)}{(2n+1)!}\zeta^{2n}\delta$ for some $\zeta\in I$.  This is less than $M^{2n}\delta/(2n)!$.
Add up all the contributions, and it is still less than $e^M\delta$.  So a $\delta$ that works for all $x\in I$ is $\epsilon e^{-M}$.  
