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I am having some problem with the following question and looking for some help with proceeding with the answer.

I am trying to prove that $$f(x)=\sum _{n=1}^{\infty} \frac {x^n}{n}$$

is continuous on the open interval (-1,1).

So plugging in terms we see that,

$$f(x)=\sum _{n=1}^{\infty} \frac {x^n}{n} = x+\frac {x^2}{2} +\frac {x^3}{3}+...$$

But how can I solve my problem?

Could I small a smaller interval $[-\beta, \beta]$ so that it converges uniformly with the smaller interval, therefore it is than continuous on the larger interval?

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    You are on the right track with the $[-\beta, \beta]$ observation because you do have uniform convergence and, therefore, continuity there. Now just take a union of such intervals as $\beta\to1$. – 2017-02-17
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    Yes, I think something like that will work. You know from the uniform continuity argument that $f(x)$ is continuous on $[-\beta,\beta]$ for any $\beta<1$ thus it is continuous on $(-1,1)$ (for any pt in $(-1,1)$ you can find an interval $[-\beta,\beta]$ with $\beta <1$ that covers it) – 2017-02-17
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    i am a little confused on where to go from here on in my workings – 2017-02-17
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    @jh123 if exists a neighborhood $U_x$ of any point $x\in(-1,1)$ such that $U_x\subseteq (-1,1)$ and $f$ converges uniformly in $U_x$ then $f$ is called **locally uniformly convergent** in $(-1,1)$. – 2017-02-17
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    could you submit a answer detailing this? – 2017-02-17

2 Answers 2

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Since the radius of convergence of the power series is $1,$ we know from the M-test that $f(x) = \sum_nx^n/n$ is uniformly convergent on $[-\beta,\beta]$ for any $\beta<1.$ Since $f(x)$ is a uniform limit of continuous functions, it is continuous on $[-\beta,\beta].$ Now recall that a function is continuous on an interval iff it is continuous on every point in that interval (that's just the definition of being continuous on a set). Take any point in $x\in(-1,1).$ We can find a $\beta<1$ such that $x\in[-\beta,\beta]$ and therefore $f$ is continuous as $x$. Thus $f$ is continuous on $(-1,1).$

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    thank you for your answer, is there a way to solve that mathematically. or is it best described with words? – 2017-02-17
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    @jh123 Well you could figure out what $\beta$ needs to be explicitly to cover $x$ and make that part more rigorous. But words are plenty mathematical when used precisely. Also see fnacool's approach, which is different, perhaps more mathematical, and correct as far as I can tell. I just wanted to post one that used your original idea to use uniform convergence. – 2017-02-17
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    how could I make that more rigorous? could you edit that in? since this was the logic I thought I should use – 2017-02-17
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    @jh123 Want to squeeze a $\beta$ in there such that $|x|<\beta<1$... Say, $\beta = (|x|+1)/2.$ You can prove that $\beta<1$, that $x\in[-\beta,\beta]$, etc. – 2017-02-17
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    how did you show using the M test that it is uniformly convergent? – 2017-02-17
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    @jh123 It's the standard proof of the fact that the power series is uniformly convergent on any interval inside its radius of convergence (which is a fact I gathered from your question that you were apprised of). I think it was what the M-test was invented for. Google any notes on power series and you'll find it. (Or just think about how to apply the M-test using the fact that you're inside the radius of convergence.) – 2017-02-17
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    I Know I need to find a series M such that f(x)<=M and M converges thus f(x) converges uniformly, but what is that series M? – 2017-02-17
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    @jh123 https://en.wikipedia.org/wiki/Weierstrass_M-test Here $f_n(x) = a_nx^n.$ So you need an $M_n$ such that $|a_n||x|^n < M_n$ and $\sum_n M_n < \infty.$ – 2017-02-17
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    @jh123 Above $|a_n||x|^n < M_n$ should be $|a_n||x|^n \le M_n$ (actually helps) – 2017-02-17
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    what is we chose 1/n? would that work? but that would be a divergent p series – 2017-02-17
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    @jh123 Not sure how $|a_n||x|^n$ and $1/n$ are even related and as you say its sum diverges anyway so no. Pick something related to $|a_n||x|^n$ that you know converges and that $|a_n||x|^n \le M_n$ for all $x\in[-\beta,\beta]$ where $\beta$ is less than the radius of convergence. – 2017-02-17
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    having trouble choosing the correct $M_n$ – 2017-02-17
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    @jh123 $M_n = |a_n| \beta^n.$ See theorem 6.2 here for more information https://www.math.ucdavis.edu/~hunter/m125a/intro_analysis_ch6.pdf – 2017-02-17
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Bare bones approach:

$$x^n -y^n = (x-y) (x^{n-1} +x^{n-2}y + \dots + y^{n-1})$$ $$\Rightarrow |x^n - y^n|\le n|x-y| \max (|x|,|y|)^{n-1}$$ (this also follows directly from the mean value theorem)

Now let $x,y\in (-1,1)$. Then

\begin{align*} |f(x) - f(y)| &=|\sum_{n=1}^\infty \frac{x^n - y^n}{n}|\\ & \le |x-y| \sum_{n=1}^\infty \max(|x|,|y|)^{n-1} \\ & = \frac{|x-y|}{1-\max(|x|,|y|)}\\ \end{align*}

Finally fix $x$ and let $y\to x$. Then denominator tends to $1-|x|$ and numerator tends to $0$. Therefore limit on RHS is $0$.

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    does this show that it is continuous on the open interval? – 2017-02-17
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    Yes. It shows that for any $x$ in the interval, $\lim_{y\to x} f(y) = f(x)$ (which is equivalent to $\lim_{y\to x} |f(y)-f(x)|=0$). – 2017-02-17