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