$\log(1-z)^{-1}$的级数$∑_{n≥1}z^n/n$收敛范围

hbghlyj

收敛半径为1. 需要证明在$|z|=1,z≠1$收敛.
因为$∑_{n≥1}z^n$在$|z|=1$不收敛,无法用函数级数积分证明(逐项积分需要一致收敛).
令$z=\cos x+i\sin x(x\in\Bbb R)$, 那么$∑_{n≥1}z^n/n=∑_{n≥1}\frac{\cos nx+i\sin nx}n$. 转化为证明$∑_{n≥1}\frac{\cos nx}n,∑_{n≥1}\frac{\sin nx}n$对$x\in\Bbb R$收敛.
因为$∑_{n≥1}\frac{\cos nx}n$不绝对收敛,不适用Weierstrass M-test.
因为$∑_{n≥1}\frac1n$和$∑_{n≥1}\cos(nx)$发散,且$f_n(x)=\cos nx$关于$n$不是单调减的,不适用Abel's test for uniform convergence.
我们使用Dirichlet's test:
级数$∑_{n≥1}\frac{\cos nx}n$在$x\in\Bbb R$不是一致收敛的，但在不含 $2k\pi,k \in \mathbb{Z}$的闭区间上一致收敛。
例如，考虑区间 $[a,\pi]$，其中 $0 < a < \pi$。我们看到 $\frac{1}{n}\searrow0$。此外，我们有
$$\left|\sum_{n=1}^m \cos nx \right| = \left|\frac{\sin \frac{mx}{2} \cos \frac{(m+1)x}{2} }{\sin \frac{x}{2}}\right| \leqslant \frac{1}{\sin \frac{a}{2} }$$
因此这些部分和对于所有 $m \in \mathbb{N}$ 和 $x \in [a,\pi]$ 都是一致有界的。
由Dirichlet's test, 级数$∑_{n≥1}\frac{\cos nx}n$在$[a,\pi]$一致收敛。

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Dirichlet's Test: Let $|\sum_{n=1}^pa_n|<K$, where $K$ is independent of $p$. Then if $f_n\ge f_{n+1}>0$ and $\lim_{n\to\infty}f_n=0$, it follows that $\sum_{n=1}^\infty a_nf_n$ converges.

hbghlyj

On the convergence of $\sum a_n \cos nx$ , $\forall x \in \mathbb R \setminus \pi \mathbb Z$ , where $\{a_n\}$ is monotone
$\sum a_n e^{inx}$ is convergent (the given sum is the real part of this). WLOG the $a_n$ are positive and decreasing. Summation by parts gives, for $x\notin 2\pi\Bbb Z$,
\begin{align*}
\bigg| \sum_{n=M}^N a_n e^{inx} \bigg| &= \bigg| \sum_{n=M}^{N-1} \bigg( \sum_{k=M}^n e^{ikx} \bigg) (a_n-a_{n+1}) + \bigg( \sum_{k=M}^N e^{ikx} \bigg) a_N \bigg| \\
&= \bigg| \sum_{n=M}^{N-1} \frac{e^{i(n+1)x}-e^{iMx}}{e^{ix}-1} (a_n-a_{n+1}) + \frac{e^{i(N+1)x}-e^{iMx}}{e^{ix}-1} a_N \bigg| \\
&\le \frac2{|e^{ix}-1|} \bigg( \sum_{n=M}^{N-1} (a_n-a_{n+1}) + a_N \bigg) = \frac2{|e^{ix}-1|}a_M.
\end{align*}
Since $a_M\to0$, this can be made less than $\varepsilon$ by choosing $M$ large enough. Therefore the series $\sum_{n=1}^\infty a_n e^{inx}$ satisfies the Cauchy criterion and hence converges.

The convergence is not necessarily uniform: $a_n=\frac1n$ is a counterexample (the resulting series is unbounded for $x$ near $2\pi\Bbb Z$).

I believe the converse is true by the following argument: if $a_n\not\to0$ then $a_n\to L$ for some $L\neq0$ (since $\{a_n\}$ is monotone); then $\sum a_n\cos nx = \sum(a_n-L)\cos nx + L \sum \cos nx$ is the sum of a series that converges for all $x\notin 2\pi\Bbb Z$ and a series that doesn't converge anywhere (since its terms don't tend to $0$), hence converges for no $x\in 2\pi\Bbb Z$. One has to deal with the case $|a_n|\to\infty$ as well....

hbghlyj

Abel's theorem

Remark

As an immediate consequence of this theorem, if $z$ is any nonzero complex number for which the series
$$\sum _{k=0}^{\infty }a_{k}z^{k}$$
converges, then it follows that
$$\lim _{t\to 1^{-}}G(tz)=\sum _{k=0}^{\infty }a_{k}z^{k}$$
in which the limit is taken from below.
The theorem can also be generalized to account for sums which diverge to infinity. If
$$\sum _{k=0}^{\infty }a_{k}=\infty $$
then
$$\lim _{z\to 1^{-}}G(z)\to \infty .$$
However, if the series is only known to be divergent, but for reasons other than diverging to infinity, then the claim of the theorem may fail: take, for example, the power series for
$${\frac {1}{1+z}}.$$
At $z=1$ the series is equal to $1-1+1-1+\cdots$, but $\frac {1}{1+1}=\frac {1}{2}$.

We also remark the theorem holds for radii of convergence other than $R=1$: let
$$G(x)=\sum _{k=0}^{\infty }a_{k}x^{k}$$
be a power series with radius of convergence $R$, and suppose the series converges at $x=R$. Then $G(x)$ is continuous from the left at $x=R$, that is,
$$\lim _{x\to R^{-}}G(x)=G(R).$$

Applications
The utility of Abel's theorem is that it allows us to find the limit of a power series as its argument (that is, $z$) approaches $1$ from below, even in cases where the radius of convergence, $R,$ of the power series is equal to $1$ and we cannot be sure whether the limit should be finite or not. See for example, the binomial series. Abel's theorem allows us to evaluate many series in closed form. For example, when
$$a_{k}={\frac {(-1)^{k}}{k+1}},$$
we obtain
$$G_{a}(z)={\frac {\ln(1+z)}{z}},\qquad 0<z<1,$$
by integrating the uniformly convergent geometric power series term by term on $[-z,0]$; thus the series
$$\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+1}}$$
converges to $\ln(2)$ by Abel's theorem. Similarly,
$$\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}$$
converges to ${\displaystyle \arctan(1)={\tfrac {\pi }{4}}.}$

$G_{a}(z)$ is called the generating function of the sequence $a.$ Abel's theorem is frequently useful in dealing with generating functions of real-valued and non-negative sequences, such as probability-generating functions. In particular, it is useful in the theory of Galton–Watson processes.

hbghlyj

Proof. We apply Dirichlet's test for convergence. The sequence of real numbers $\left\{\frac{1}{n}\right\}$ is non-increasing and converges to 0. By the triangle inequality,
\[
\left|\sum_{n=1}^N z^n\right|=\left|\frac{z-z^{N+1}}{1-z}\right| \leq \frac{|z|+|z|^{N+1}}{|1-z|}=\frac{2}{|1-z|}
\]
Since $\frac{2}{|1-z|}$ is bounded for fixed $z \neq 1$ on the unit circle, $\sum_{n=1}^{\infty} \frac{z^n}{n}$ converges.