Prove that $\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{2k+1}x^{2k+1}$ uniformly converges on $[-1,1].$

Question

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?

Answer 1

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:

1. $\sum_{k \ge 1} (-1)^k a_k$ converges to some $s$
2. $|s - \sum_{1 \le k \le n}(-1)^k a_k|<a_{n+1}$

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 $$|f(x)-f_n(x)|\le \frac{ |x|^{2n+3} }{2n+3}\le \frac{1}{2n+3} \to 0$$

Answer 2

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:

$$|f_n(x)-f(x)|=\left| \int_0^x \frac{s^{2(n+1)}}{1+s^2}\,ds\right| \le \left| \int_0^1 s^{2(n+1)}\,ds\right|=\frac{1}{2n+3}\to0$$