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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.

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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 $x_2$.

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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 $\epsilon e^{-M}$.

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  • $\begingroup$ This is a solid answer, so well done! But I believe that in the post it states that the OP is seeking a way forward that involves applying theorems such as the Weierstrass M-Test. I like your solution in that it is fundamentally based. Yet, I am not sure that this is exactly what the OP requested. ;-) $\endgroup$ – Mark Viola Oct 21 '15 at 15:32
  • $\begingroup$ @Michael What is the motivation behind having $|\sin(x) - \sin(x + \delta)| < \epsilon?$. Isn't uniform convergence about choosing an $N$ such that $n \geq N$ implies $|f_n(x) - f(x)| < \epsilon$? $\endgroup$ – user247618 Oct 21 '15 at 15:56
  • $\begingroup$ Hmm, sorry, I seem to have shown it was uniformly continuous instead. $\endgroup$ – Empy2 Oct 21 '15 at 16:57

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