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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 $a\in(-R,R)$

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  • $\begingroup$ It seems you're missing a "$\sum$". $\endgroup$ Apr 21, 2019 at 7:13
  • $\begingroup$ @David Yes, you are correct. I will add it $\endgroup$
    – Anvit
    Apr 21, 2019 at 7:13
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    $\begingroup$ You have $a\in(-1,1)$, so it can't be $-1$. What does divergence at $-1$ have to do with anything? $\endgroup$ Apr 21, 2019 at 7:15
  • $\begingroup$ @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 $\endgroup$
    – Anvit
    Apr 21, 2019 at 7:20
  • $\begingroup$ 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 $\endgroup$
    – Henry
    Apr 21, 2019 at 7:40

2 Answers 2

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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.

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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 $|a_nb_n+\cdots+a_mb_m|<Mb_0$

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