Prove that $\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{2k+1}x^{2k+1}$ uniformly converges on $[-1,1].$ Prove that $\sum\limits_{k=1}^{\infty}\dfrac{(-1)^k}{2k+1}x^{2k+1}$ uniformly converges on $[-1,1].$
My book says I have use alternating series test. I can see that the series converges for any $x\in[-1,1]$ by the alternating series test but it doesn't tell us the series is uniformly convergent on $[-1,1]$. I tried to use Weierstrass M-Test instead but it fails to pass the M-test. 
I let $f_k(x) = \dfrac{(-1)^k}{2k+1}x^{2k+1}$ and found $M_k = \sup\{|\dfrac{(-1)^k}{2k+1}x^{2k+1}|:x\in[-1,1]\} = \dfrac{1}{2k+1}$. But $\sum\limits_{k=1}^{\infty}\dfrac{1}{2k+1}$ is not convergent and does not pass the M-test. Hence, the series cannot be uniformly convergent. 
Did I use the M-Test correctly?  

Let $$f_n(x)=\sum_{k=1}^{n} \frac{(-1)^k}{2k+1} x^{2k+1},$$ and call $$f(x)=\sum_{k=1}^{\infty} \frac{(-1)^k}{2k+1} x^{2k+1}.$$ 
Note that $$f_n'(s)=-s^2\frac{1-(-s^2)^n}{1+s^2} \implies f(x)=- \int_0^x s^2\frac{1-(-s^2)^n}{1+s^2}\,ds$$ and $$f(x)=\arctan(x)-x=-\int_0^x\frac{s^2}{1+s^2}\,ds.$$
Thus:
$$|f_n(x)-f(x)|=\left| \int_0^x \frac{s^{2(n+1)}}{1+s^2}\,ds\right| \le \left| \int_0^1 s^{2(n+1)}\,ds\right|=\frac{1}{2n+3}\to0$$
