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In calculus:

Given $\displaystyle \sum_{n=0}^{\infty} {\frac{n^2 2^n}{n^2 + 1}x^n} $, prove that it converges for $-\frac{1}{2} < x < \frac{1}{2} $, and that it does not converge uniformly in the area of convergence.

So I said:

$R$ = The radius of convergence $\displaystyle= \lim_{n \to \infty} \left|\frac{C_n}{C_{n+1}} \right| = \frac{1}{2}$ so the series converges $\forall x$ : $-\frac{1}{2} < x < \frac{1}{2}$.

But how do I exactly prove that it does not converge uniformly there? I read in a paper of University of Kansas that states: "Theorem: A power series converges uniformly to its limit in the interval of convergence." and they proved it.

I'll be happy to get a direction.

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    $\begingroup$ If you are referring to this paper (there's no author and no title, hence it's not really a paper), then the thoerem is clearly wrong. The proof given there is correct, but for a different theorem: For all $\rho<r$, the convergence on $[-\rho,\rho]$ is uniform. Note that this is something different than saying that the convergendce on $(-r,r)$ is uniform! Don't trust some weird folks making mathematical claims on the internet! (Unless it is us, that is) $\endgroup$ – Hagen von Eitzen May 18 '13 at 10:18
  • $\begingroup$ ... But maybe reading the proof there (which, judging by the problem statement, you should have encountered in your lecture just a few hours ago) can help you find out how things can go wrong with the series in the problem statement and thus how to show that here the convergence is not uniform. $\endgroup$ – Hagen von Eitzen May 18 '13 at 10:22
  • $\begingroup$ The terms $f_n(x)={n^2 2^n\over n^2+1}x^n$ do not converge uniformly to $0$ on $(-1/2,1/2)$ (to see this, compute $\lim_{x\rightarrow 1/2^-} f_n(x)$). Consequently, the series does not converge uniformly on $(-1/2,1/2)$. $\endgroup$ – David Mitra May 18 '13 at 10:43
  • $\begingroup$ You mean calculating the limit when $n \to \infty$ and $x \to \frac{1}{2}$? If so, the result is $1$ $\endgroup$ – TheNotMe May 18 '13 at 10:48
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    $\begingroup$ No. I'm appealing to the fact that if $\sum f_n(x)$ converges uniformly on $I$, then given $\epsilon>0$, there is an $N$ so that for $n>N$, one has $|f_n(x)|<\epsilon$ for all $x\in I$ (that is, the terms $f_n$ converge uniformly to $0$ on $I$. But here, for any $n$, one has $\lim_{x\rightarrow 1/2^-} {n^2 2^n\over n^2+1}x^n={n^2\over n^2+1}>1/2$. So, $( {n^2 2^n\over n^2+1}x^n)$ does not converge uniformly to $0$ on $(-1/2,1/2)$. $\endgroup$ – David Mitra May 18 '13 at 11:54
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Let's prove a more general statement: if the coefficients of power series $\sum a_n x^n$ are such that the limit
$$\lim_{n\to\infty} \frac{a_n}{R^n} $$ is finite and nonzero, then

  1. the interval of convergence is $(-R,R)$
  2. the series does not converge uniformly on $(-R,R)$

The first part is immediate from ratio test and the fact that $a_nx^n\not\to 0$ when $|x|\ge R$. For the second part, use the fact that for every $n$ $$\sup_{|x|< R} |a_n x^n| = \sup_{|x|\le R} |a_n x^n| = |a_n| R^n$$ which by our assumption does not approach zero. Therefore, the uniform convergence fails: even individual terms of the series are not uniformly small, let alone the remainder.

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According to Hagen von Eitzen theorem, you know that it will uniformly (and even absolutly) converge for intervals included in ]-1/2;1/2[. So the problem must be there.

The definition of uniform convergence is : $ \forall \epsilon>0,\exists N_ \varepsilon \in N,\forall n \in N,\quad [ n \ge N_ \varepsilon \Rightarrow \forall x \in A,d(f_n(x),f(x)) \le \varepsilon] $

So you just have to prove that there is an $\varepsilon$ (1/3 for instance or any other) where for any $N_ \varepsilon$ there will always be an $x$ (look one close from 1/2) where $d(f_n(x),f(x)) \ge \varepsilon $ for an $n \ge N_ \varepsilon$

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  • $\begingroup$ but, $f(x)$ here does not exist, because $\lim_{n \to \infty} f_n(x)$ is $\infty$ $\endgroup$ – TheNotMe May 18 '13 at 12:05
  • $\begingroup$ No, it's 0 for x<1/2, 1 for x=1/2, and infinite otherwise. So it exists on [-1/2;1/2] $\endgroup$ – gvo May 18 '13 at 12:24
  • $\begingroup$ I'm also wondering if you couldn't use the fact that f isn't continuous to prove that it does not converge uniformly. I think it's true for the sequence of fn but I don't know for the series. $\endgroup$ – gvo May 18 '13 at 12:33

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