# 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?

• @πr8 Oops, I thought $x^k$, not $x^{2k+1}$ ... – Hagen von Eitzen Apr 19 '17 at 20:19
• The Weierstrass $M$-test isn't a “test” in the way that the Ratio or Root Test is. You can't use it to conclude that a power series does not converge uniformly. – Matthew Leingang Apr 19 '17 at 20:31
• @MatthewLeingang Then what would be the purpose of using the M-test then? I am really confused because I've been using it to test convergence for series problems. Are you saying if M-test fails, I can't conclude anything about the series? – user3000482 Apr 20 '17 at 0:27
• By "if the $M$-test fails", I guess you are referring to the fact that $\sum_{k=1}^\infty \sup\left\{\left|\frac{(-1)^kx^{2k+1}}{2k+1}\right| : x \in [-1,1]\right\}$ does not converge. That's not a failure of the test; rather, the test does not apply. It just means you have to use another method. – Matthew Leingang Apr 20 '17 at 0:54

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

• But if we use the alternating series test, don't we only know that the series is convergent not uniformly convergent? – user3000482 Apr 20 '17 at 21:30
• You have a uniform bound on $|f(x)-f_n(x)|$; this is what is needed for uniform convergence. – πr8 Apr 20 '17 at 22:48
• I'm sorry but I am not sure what you mean by uniform bound. Don't we need to make sure that $\| f(x)-f_n(x) \|_{\sup} < \epsilon$? – user3000482 Apr 21 '17 at 15:05
• A uniform bound is the same as a supremum bound, yes. $\| f(x)-f_n(x) \|_{\sup} = \sup_{x\in [-1,1]} |f(x)-f_n(x)|$. Since $|f(x)-f_n(x)|<\frac{1}{2n+3}$ for all relevant $x$, it is clear that $\sup_{x\in [-1,1]} |f(x)-f_n(x)|<\frac{1}{2n+3}$ also. So yes, the series certainly is uniformly convergent. – πr8 Apr 21 '17 at 15:23
• Maybe - one could conceivably adapt the proof above to prove a result of this form for power series with decaying coefficients of alternating sign. – πr8 Apr 21 '17 at 18:35

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:

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

• I'm sorry but I dont' quite understand your answer. How can you take derivative of $f_n(x)$? And what is the variable $s$? Is it just an arbitrary element from $[-1,1]$? I feel frustrated because I don't know how am I supposed to come up with such crazy proof... if that's the only way to solve this. Can you comment on my attempt of using M-test? I think I used it properly but the series fails the M-test. – user3000482 Apr 19 '17 at 20:22
• 1) You can take the derivative of $f_n$ by differentiating it term-by-term; $f_n$ is just a polynomial. 2) Yes, $s$ is just a dummy variable. 3) To show uniform convergence, you need to control the difference between the partial sums ($f_n$ in this question) and the true limit ($f$), and integrals often end up being a good way of doing so. 4) The M-test doesn't work here because, as you say, the terms in the resulting sum do not have a convergent sum. – πr8 Apr 19 '17 at 20:25