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 $[-1, 1]$

  • $\begingroup$ $g(x)$ diverges as $x \to -1$. $\endgroup$ – marty cohen Feb 24 at 4:34
  • 2
    $\begingroup$ @martycohen No it doesn't. The series is still alternating because the exponents all have the same parity. $\endgroup$ – Jyrki Lahtonen Feb 24 at 4:37
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Feb 24 at 4:42
  • $\begingroup$ Oops - you are right. I read the exponent as k+1 rather than 2k+1. $\endgroup$ – marty cohen Feb 24 at 6:48

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

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