How do I show that the Taylor series of $\ln(x)$ at point $x=1$ doesn't converge uniformly on any interval $(0,\delta)\subseteq(0,2)$?
My edited approach:
Let be $S_n(x):= \sum\limits_{k=1}^n \frac{(-1)^{k+1}}k(x-1)^k$ the $n$-th Taylor polynomial of $\ln(x)$ at point $x=1$.
The ieda is to prove the statement by showing that the Cauchy-criterion can't be satisfied, namely:
$\exists \epsilon>0$ such that for all $n_0\in\mathbb{N}$ there are two $n>m>n_0$ such that $\Vert S_m-S_n\Vert_{\infty}\geq\epsilon$.
First we assume $0<\delta\leq1$ and define a sequence by $x_n:=1-(\frac{1-\delta}{2})^{\frac{1}{n}}$, clearly $x_n\in(0,\delta)$.
We observe that: $$\Vert S_m-S_n\Vert_{\infty}\geq|S_n(x_n)-S_m(x_n)|=|\sum\limits_{k=m+1}^n\frac{(-1)^{k+1}}k(x_n-1)^k|=|\sum\limits_{k=m+1}^n\frac{(-1)^{2k+1}}k(1-x_n)^k|\\=|-\sum\limits_{k=m+1}^n\frac{(1-x_n)^k}k|\geq \sum\limits_{k=m+1}^n\frac{(1-1+(\frac{1-\delta}{2})^{\frac{1}{n}})^n}k=\frac{1-\delta}{2}\sum\limits_{k=m+1}^n\frac{1}k.$$ As the harmonic series doesn't converge at all the Cauchy criterion can't be satisfied.
However, the case when $1<\delta<2$ doesn't seem so easy. At first glance it looks like that the Taylor series could converge uniformly because there appears an alternating series and $(x-1)^k$ resembles a geometric series!?
$|S_n(x)-S_m(x)|=|\sum\limits_{k=m+1}^n\frac{(-1)^{k+1}}k(x-1)^k|=…$?
Is the first part o.k.? How do I proceed in the second part?
Best Answer
First consider uniform convergence on any interval $(0,\delta)$ where $\delta \leqslant1$.
We have
$$\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}k(x-1)^k\right|= \sum_{k=n+1}^{\infty}\frac{(1-x)^k}k \geqslant\sum_{k=n+1}^{2n}\frac{(1-x)^k}k ,$$
and
$$\sup_{x \in (0,\delta)}\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}k(x-1)^k\right|\geqslant \sup_{x \in (0,\delta)}\sum_{k=n+1}^{2n}\frac{(1-x)^k}{k} \geqslant \sum_{k=n+1}^{2n}\frac{(1-1/n)^k}{k}\\\geqslant n \cdot \frac{(1-1/n)^{2n}}{2n} = \frac{(1-1/n)^{2n}}{2}\underset{n \to \infty}\longrightarrow \frac{e^{-2}}{2}\neq 0$$
Hence, the convergence is not uniform on $(0,\delta)$ with $\delta \leqslant 1$. Of course, that means the series fails to converge uniformly on any interval $(0,\delta)$ with $\delta <2$, since the source of the non-uniform convergence is the behavior near the endpoint $x = 0$.
On the interval, $[\delta, 1]$ where $\delta < 1$, the series takes the form $-\sum_{k=1}^{\infty}\frac{(1-x)^k}{k}$. Since, $\frac{(1-x)^k}{k} \leqslant (1-\delta)^k$ we have uniform convergence here by the Weierstrass M-test.
You should be able to show that convergence is also uniform on the interval $[1,2]$. Hint: Dirichlet test.