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Prove that the series $\sum_{k=1}^{\infty} \frac {x^k} {k}$ does not converge uniformly on the interval $[0, 1)$.

Thoughts: In order to show this I am not sure where to begin. Do I suppose that $\sum_{k=1}^{\infty} \frac {x^k} {k}$ does in fact converge and then find some contradiction, or do I show the negation of the definition of uniform convergence is true. Any hints much appreciated.

Kernel_Dirichlet
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IntegrateThis
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    if it converged uniformly, the limit would have to be bounded – user8268 Apr 16 '17 at 06:18
  • @user8268 the limit being the harmonic series? – IntegrateThis Apr 16 '17 at 06:19
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    sorry for being unclear - I meant that the function $\sum_{k=1}^\infty x^k/k$ on $[0,1)$ would have to be bounded. But you are right that one can see that it's not bounded by comparing with the harmonic series. But perhaps you know what that function is. – user8268 Apr 16 '17 at 06:24
  • @user8268 i understand what you are saying intuitively but I have no idea how to make this more rigorous using the definition. – IntegrateThis Apr 16 '17 at 06:32
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    try proving: in $f_n\to f$ uniformly and if each $f_n$ is a bounded function then $f$ is bounded (easy). Then apply in your situation (either knowing that the sum is $-\log(1-x)$ or by noticing that for any $N$ you can find an $x$ s.t. $\sum_{k=1}^\infty x^k/k>\frac{1}{2}\sum_{k=1}^N 1/k$, or perhaps you'll find another method) – user8268 Apr 16 '17 at 06:38
  • Closely related. So closely that IMHO this could be closed as a duplicate. Because I answered the target question I will refrain from voting. – Jyrki Lahtonen Apr 16 '17 at 07:49

1 Answers1

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We know the series converges pointwise, using, for example, the ratio test. Showing that the negation of the definition for uniform convergence is true is facilitated by the following argument.

Let

$$S(x) = \sum_{k=1}^\infty \frac{x^k}{k}.$$

Uniform convergence requires for every $\epsilon > 0$ there exists an integer $N(\epsilon)$ independent of $x$ such that if $n \geqslant N(\epsilon)$ then for every $x \in [0,1)$ we have

$$\left|\sum_{k=1}^n \frac{x^k}{k} - S(x)\right| = \sum_{k = n+1}^ \infty \frac{x^k}{k} < \epsilon,$$

which implies, for all $n \geqslant N(\epsilon),$

$$\sup_{x \in [0,1)}\sum_{k = n+1}^\infty \frac{x^k}{k} < \epsilon,$$

or, equivalently,

$$\lim_{n \to \infty}\sup_{x \in [0,1)}\sum_{k = n+1}^\infty \frac{x^k}{k}= 0$$

However, for all $n$,

$$\sup_{x \in [0,1)} \sum_{k=n+1}^\infty \frac{x^k}{k} \geqslant \sup_{x \in [0,1)} \sum_{k=n+1}^{2n} \frac{x^k}{k} \geqslant \sup_{x \in [0,1)} n \frac{1}{2n}x^{2n} = \frac{1}{2},$$

and

$$\lim_{n \to \infty}\sup_{x \in [0,1)}\sum_{k = n+1}^\infty \frac{x^k}{k} \neq 0.$$

Thus, the series fails to converge uniformly on $[0,1)$.

RRL
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