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?  

Note that for $x\in[0,1]$, the sequence $\frac{x^{2k+1}}{2k+1}$ is monotone decreasing (and for $x\in[-1,0]$, the negative of this sequence is monotone decreasing).
If $a_k$ is a monotone decreasing, nonnegative sequence, the alternating series test says that:


*

*$\sum_{k \ge 1} (-1)^k a_k$ converges to some $s$

*$|s - \sum_{1 \le k \le n}(-1)^k a_k|<a_{n+1}$


Thus, if $f_n(x)=\sum_{k=1}^{n} \frac{(-1)^k}{2k+1} x^{2k+1}$ and $f(x)=\sum_{k=1}^{\infty} \frac{(-1)^k}{2k+1} x^{2k+1}$, then $$|f(x)-f_n(x)|\le \frac{ |x|^{2n+3} }{2n+3}\le \frac{1}{2n+3} \to 0$$
