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I am studying sequences and series of functions and in the course notes there is this excercise:

Prove the function $$f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$ has second derivative.

So i started proving it is derivable in the following way:

$f'(x)=\lim_{h \to 0} \frac{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(x+h)^{2n+1} - \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}}{h}$

$f'(x)=\lim_{h \to 0} \lim_{n \to \infty} \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!} \frac{(x+h)^{2k+1}-x^{2k+1}}{h}$

Now, if i could interchange those limits, i mean if i could write: $$f'(x)=\lim_{n \to \infty} \lim_{h \to 0} \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!} \frac{(x+h)^{2k+1}-x^{2k+1}}{h}$$ it would follow easily from the derivability of the identity function. But i'm not sure how i can do this, it seems to me that i need to prove the functions $$g_n(h)= \sum_{k=0}^n \frac{(-1)^k}{(2k+1)!} \frac{(x+h)^{2k+1}-x^{2k+1}}{h}$$ converges uniformely to be able to make that switch of limits and that seems to reduce to the problem of proving that $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$ converges uniformely. Is this procedure right, and if that case how can i prove that? i can't use Weierstrass M-test here since $\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ is not bounded above. Any advice is welcome

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The Taylor series for $\sin (x)$ is

$\sin (x) = \sum_{0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$

The series converges uniformly on any bounded subset of $\mathbb R$.

You can use M-test with $M_n = \frac{R^{2n+1}}{(2n+1)!}$ for the interval $[-R,R]$ ($R\in \mathbb R$)

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The coefficients of such power series decay so fast that they ensure that $f(x)$ is an analytic function with radius of convergence $+\infty$. We are so allowed to differentiate $f(x)$ termwise, even twice, and check that $f''(x)=-f(x)$. Since $f(0)=0$ and $f'(0)=1$, $f(x)=\sin x$ for every $x$, because the previous ODE has a unique global solution.

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