# Prove $g(x) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\,x^{2k+1}$ converges uniformly on [-1,1]

For the problem below, can someone please tell me how to enhance my proofs? I am not confident in my proofs. Thank you!

$$\textbf{Problem:}$$Let $$\displaystyle{g(x) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\,x^{2k+1}}$$.

i. Prove the sum defining $$g(x)$$ converges uniformly on $$[-1,1]$$.

$$\textbf{Proof:}$$ Since $$x^{2k+1}$$ is positive, monotonic, and bounded for all $$x\in [-1,1]$$, and the series $$\displaystyle{\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}}$$ is uniformly convergent series because $$\displaystyle{\sum_{k=1}^\infty \frac{(-1)^k}{2k+1}}$$ is alternating series and $$\frac{1}{2k+1}$$ is decreasing and tends to 0. So, by Leibniz test $$\displaystyle{\sum_{k=1}^\infty \frac{(-1)^k}{2k+1}}$$ is convergent series.

Therefore, $$\displaystyle{\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} x^{2k+1}}$$ is converging uniformly on $$[-1,-1]$$. Hence, $$g(x) = \displaystyle{\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} x^{2k+1}}$$ is converging uniformly.

ii. Prove $$g\in{\mathcal C}[-1,1]$$, that is, that $$g$$ is continuous on $$[-1,1]$$.

$$\textbf{Proof:}$$ We know if a series $$\sum f_n$$ converges uniformly to $$f$$ in an interval $$[a, b]$$ and it's terms $$f_n$$ are continuous at a point $$x_0 \in [a,b]$$, then some function $$f$$ is also continuous at $$x_0$$. Here each term of the series is continuous and series is uniformly convergent and converge to $$g(n)$$ so $$g(n)$$ is continuous on $$[-1, 1]$$

• $g(x)$ diverges as $x \to -1$. – marty cohen Feb 24 at 4:34
• @martycohen No it doesn't. The series is still alternating because the exponents all have the same parity. – Jyrki Lahtonen Feb 24 at 4:37
• I think the argument is fine. You can end the first proof a bit sooner, I think. After all, Leibniz theorem on alternating series also gives you a bound on the cut-off error. Here that bound is independent of $x$, giving us uniform convergence right away. I woud skip Abel altogether. Anyway, uniform convergence preserves continuity. As described in your part two. – Jyrki Lahtonen Feb 24 at 4:42
• Oops - you are right. I read the exponent as k+1 rather than 2k+1. – marty cohen Feb 24 at 6:48

Here is another attempt: $$\left| {\sum\limits_{k = 0}^\infty {\frac{{( - 1)^n }}{{2k + 1}}x^{2k + 1} } - \sum\limits_{k = 0}^{n - 1} {\frac{{( - 1)^n }}{{2k + 1}}x^{2k + 1} } } \right| = \left| {\int_0^x {\frac{{t^{2n} }}{{1 + t^2 }}dt} } \right| \le \int_0^{\left| x \right|} {\frac{{t^{2n} }}{{1 + t^2 }}dt} \le \int_0^1 {t^{2n} dt} = \frac{1}{{2n + 1}}.$$