Uniform Convergence of $\sum_1^\infty(-x)^n/n$ on $[a, 1]$

How can I prove that $$\sum_1^\infty(-x)^n/n$$ converges uniformly on $$[a, 1]$$ where $$a\in(-1,1)$$

I've tried to use both Cauchy's criteria and Weierstrass M Test but have failed since the series does not converge at $$-1$$. I still think Cauchy criteria might work but I'm unable to find the right inquality I guess

In general, how can I prove that if a power series converges on $$(-R,R]$$ then in converges uniformly on $$[a,R]$$ where $$a\in(-R,R)$$

• It seems you're missing a "$\sum$". Apr 21, 2019 at 7:13
• @David Yes, you are correct. I will add it Apr 21, 2019 at 7:13
• You have $a\in(-1,1)$, so it can't be $-1$. What does divergence at $-1$ have to do with anything? Apr 21, 2019 at 7:15
• @DavidMitra When applying Cauchy's criteria (and M Test), the absolute values is causing problems. If it did converge at $-1$ as well, I would be done Apr 21, 2019 at 7:20
• If $0 \le a \lt 1$ then it is not difficult to find a bound for uniform convergence from $\sum_1^\infty(-1)^n/n$. So if $-1\lt a \lt 0$ you could consider $x \in [|a|,1]$ and $x \in [-|a|,|a|]$ separately to find a uniform convergence bound for each, and then choose the higher one Apr 21, 2019 at 7:40

This follows from Dirichlet's test, since $$\left(\frac1n\right)_{n\in\mathbb N}$$ is monotonic and converges to $$0$$ and the partial sums of $$\sum_{n=0}^\infty(-x)^n$$ are uniformly bounded.
Apply Abel's Lemma to the sequences $$(a_nR_n)$$ and $$(x/R)^n$$ to get for all $$n > m≥N$$ and all $$x\in[0,R]$$ that $$∣a_mR_m(x/R)^m+···+a_nR_n(x/R)^n∣\le \epsilon(x/R)^m < \epsilon$$
Abel's Lemma States that if $$b_1\ge b_2\ge\dots$$ and $$\sum_1^m a_n$$ is bounded (for all $$m$$) (by say M) then $$|a_nb_n+\cdots+a_mb_m|