uniform convergence of series of functions $\sum_1^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$ The series of functions given is 

$$\sum_1^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$$

is uniformly convergent on any closed and bounded interval $[a,b]$
solution I tried-we know that $$\sum_0^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}=\cos x$$
so here the given series is $$1+\sum_1^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}=\cos x$$
$$\Rightarrow \sum_1^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}=\cos x -1$$
what can I say about convergence?
Is there any another method to prove its uniform convergence, these questions are very tough to solve 
Please help
 A: Some hints:
I am sure you noticed that $$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$ is a power series. A power series converges uniformly in any closed interval 
$[a,b] \subset (-R,R), R $ here being the radius of convergence. Not sure if you are familiar with this theorem, it is an immediate consequence of Weierstrass M-test, so convince yourself that it works before you use it!. So to conclude your proof, you may continue this way: 


*

*find the radius of convergence of your power series (you already know it converges to $cos(x)$, which basically gives you the answer immediately)

*Prove and use the theorem I mentioned

A: In the interval $[a, b]$ it is true that $\lvert x \rvert^{2n} \leq (\max(\lvert a \rvert, \lvert b \rvert))^{2n}$. Hence, 
$$\bigg \lvert \sum_{n = 1}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \bigg \rvert \leq \sum_{n = 1}^\infty \frac{(\max(\lvert a \rvert, \lvert b \rvert))^{2n}}{(2n)!}.$$
The series on the right hand side is convergent by the ratio test. Therefore, by the Weierstraß M-Test, the series on the left hand side converges uniformly on the interval $[a, b]$.
