Proving $\sum_{n=1}^\infty \frac{\xi ^n}{n}$ is not uniformly convergent Proving $\sum_{n=1}^\infty \frac{\xi ^n}{n}$ is not uniformly  convergent for $\xi \in (0,1)$.
I am trying to do the above. I have attempted to show it is not a cauchy sequence by considering $||\frac{\xi ^n}{n} ||_{\sup}$ but no avail. Any help please
 A: Hint: Show for $z$ close to $1$ that the convergence of the series becomes arbitrarily slow. More formally, show the negation of uniform convergence: There exists $\epsilon > 0$ such that for all positive integers $N$, there exists $z \in (0,1)$ and $n \geq N$ such that $|f_n(z) - f(z)| \geq \epsilon$. Here $f_n$ is the $n$th partial sum of your series, and $f$ is the limiting function. It should be clear that this is true because you can choose $z$ to be whatever you want, arbitrarily close to $1$.
A: The series fails to converge uniformly if you can show there is some $\epsilon_0 > 0$ such that for every $N \in \mathbb{N}$,
$$\sup_{\xi \in (0,1)} \sum_{n=N}^\infty\frac{\xi^n}{n}\geqslant \epsilon_0.$$
Notice that
$$\sup_{\xi \in (0,1)} \sum_{n=N}^\infty\frac{\xi^n}{n}\geqslant \sup_{\xi \in (0,1)} \sum_{n=N+1}^{2N}\frac{\xi^n}{n}\geqslant \sup_{\xi \in (0,1)}N\frac{\xi^{N+1}}{2N} = \frac{1}{2}.$$
Hence, using $\epsilon_0 = 1/2$ confirms the negation and the series is not uniformly convergent on $(0,1)$.
A: Let $f_n$ be the $n$th partial sum function, and consider $|f_{2n} - f_n|_1$. This is equal to
$$H_{2n} - H_n$$
where $H_n = \sum_{k=1}^n 1/k $ is the $n$th harmonic series partial sum. As $n \to \infty,$ we have $H_n \to \ln n + \lambda$ for some constant $\lambda$ independent of $n$ (look up Euler constant if you haven't seen this before). Then
$$H_{2n} - H_n \simeq \ln 2n - \ln n = \ln 2$$.
