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I know that $f(x)=\sum_{n=1}^{\infty}\frac{x^n}{\sqrt{n}}$ converges for every $x \in [-1,1)$ but not for $x=1$. How can I use Weierstrass M-test to show that $f$ is continuous at a fixed $x_0 \in (-1,1)$?

uniquesolution
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Maria Fernandez
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  • Do you really want to use the $M$ test? I mean, the series is absolutely converging and $x^n/\sqrt{n}$ is continuous near $0$. – Nicolas Dec 17 '20 at 20:21
  • The series converges uniformly for $x$ on any compact subinterval of $(-1,1)$. Hence, $f(x)$ is continuous on all such intervals. Hence, $f$ is continuous on $(-1,1)$. – Mark Viola Dec 17 '20 at 20:24
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    Alternatively, we have $$\left|\sum_{n=N}^\infty\frac{x^n-x_0^n}{\sqrt{n}}\right|\le \frac1{\sqrt{N}}\left|\frac{x^{N+1}}{1-x}-\frac{x_0^{N+1}}{1-x_0}\right|$$which for any $x\in (-1,1)$ and $x_0\in (-1,1)$ goes to $0$ as $N\to\infty$ – Mark Viola Dec 17 '20 at 20:33

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Let $x_0\in(-1,1)$. If you find some interval around $x_0$ where the series above converges uniformly, then the limit must be continuous (as the uniform limit of continuous functions). To show uniform convergence, you can use the Weierstrass M-test - choose your small interval, and then find a uniform bound on the series in that interval.

Since $x_0\in (-1,1)$, there is some $\epsilon>0$ such that $(x_0-\epsilon,x_0+\epsilon)\subset(-1,1)$. Choose this to be your interval. Let $M=max\{|x_0+\epsilon|,|x_0-\epsilon|\}$. Then by comparing your series to the series $\sum_{n=1}^{\infty}\frac{M^n}{\sqrt{n}}$ (and recalling that $M<1$), you see by the Weierstrass M-test that your series of functions converges uniformly in the interval you chose, which means that the limit function of the series is continuous at $x_0$.

Note that you can use other arguments which are a bit simpler by using the notion of radius of convergence, but if you want to use Weierstrass, this is a way.

GSofer
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