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I am looking for some guidance with the following questions. I believe I have done question 1) correctly, but question 2) which is the more general case I am not sure how to prove this

1) Given the function,

$$f(x)=\sum _{n=1}^{\infty}\frac{x^n}{n^2}$$ is continuous on the closed interval $[-1,1]$

By the alternating series test this series converges at $x=-1$ and looking at $x = 1$ we see that this is simply just a convergent p series with $p=2>1$.

Now considering the interval $-1\leq a\leq 1$ and taking into consideration that $\sum\frac1{n^2}a^n$ converges. Applying the Weierstrass M-test, since $|n^{-2}x^n|\geq|n^{-2}a^n| = \left(\frac{a^n}{n^2}\right)$ for $x \in [-a,a]$, then $\sum\frac1{n^2}x^n$ converges uniformly to a function on $[-a, a]$. Since $|a|$ can be any number $\leq1$, therefore $f$ represents a continuous function on $[-1, 1]$.

2) However with this more general case I am having problem with its proof. If the series $\sum a_n$ converges then $$f(x)= \sum _{n=0} ^{\infty}a_nx^n$$ defines a continuous function on $(-1,1)$

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  • $\begingroup$ I added "converges" for the series in 2). But you should clarify if you mean that it converges absolutely. $\endgroup$ – uniquesolution Mar 8 '17 at 20:41
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A fact about power series is that there is a (potentially zero or infinite) number $R$ with the property that $\displaystyle \sum_{n=0}^\infty a_n x^n$ converges uniformly on every compact subinterval of $(-R,R)$ and diverges whenever $|x| > R$. According to the fact $\displaystyle \sum_{n=0}^\infty a_n$ converges you have $R \ge 1$. Thus the series converges uniformly (hence to a continuous function) on every compact subinterval of $(-1,1)$.

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