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Prove the uniform convergence of the functional series in the indicated interval $$\sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{\sqrt{n}}\ \ \ \ \ \ x \in [0,1]$$

I tried the M-test and got $|(-1)^{n-1} \frac{x^n}{\sqrt{n}}|=|\frac{x^n}{\sqrt{n}}| \leq x^n$, and since $\sum_{n=1}^\infty x^n$ converges for $|x|<1$, and so the original series converges uniformly for $x \in [0,1)$. But this way the interval doesn't include $1$.

Can anyone give me a better way to prove this? Thanks in advance.

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  • $\begingroup$ $$\lim_{m\to\infty}\sum_{n=1}^{m}\frac{(-1)^{n-1}x^n}{\sqrt{n}}=$$ $$\lim_{m\to\infty}\left((-x)^{m+1}\Phi\left(-x,\frac{1}{2},m+1\right)-\text{Li}_{\frac{1}{2}}(-x)\right)=$$ $$\text{Li}_{\frac{1}{2}}(-x), |x|<1$$ $\endgroup$ – Jan Nov 22 '15 at 17:58
  • $\begingroup$ I'd like Lucy to show how she showed uniform convergence on [0,1) since adding 1 more point to the domain and retaining uniformity of convergence is nearly trivial. $\endgroup$ – DanielWainfleet Nov 22 '15 at 18:04
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    $\begingroup$ I don't think you applied the $M$ test correctly, you have to obtain a series independent from $x$. $\endgroup$ – mrprottolo Nov 22 '15 at 18:09
  • $\begingroup$ Even if your previous work was wrong, you might want to leave it so that this question does not get closed in review for not having proper context. Then it risks getting deleted. $\endgroup$ – robjohn Nov 22 '15 at 18:37
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Write $R_N(x)=\sum_{n\ge N} (-1)^{n-1}x^n/\sqrt{n}$. Then $|R_N(x)|\le 1/\sqrt{N}$ for $0\le x\le 1$ because the series is alternating and the terms decrease in absolute value (this is the argument from the proof of the Leibniz criterion).

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