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.
Best Answer
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 $x_2$.