我如何将我的问题编号为 1、3 和 14？以下代码：
\documentclass{article}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{graphicx}
\usepackage{setspace}
\usepackage{amsfonts}
\onehalfspacing
\DeclareMathOperator{\Log}{Log}
\DeclareMathOperator{\Arg}{Arg}
\begin{document}
\title{Homework Chapter 5}
\author{}
\maketitle
\setcounter{section}{5}
\setcounter{subsection}{1}
\subsection{Problems 1, 3, and 14}
\begin{enumerate}[wide=0pt, label=\bfseries\arabic*., start=1]
\item (a) Prove that the power series $\sum_{j=0}^{\infty} z^{j}$ converges at no point on its circle of convergence $|z|=1$.
Let us first prove that if the sequence $\{z_{n}\}_{n=1}^{\infty}$ converges, then $(z_n - z_{n-1})\rightarrow 0$ as $n \rightarrow \infty$.
Proof. Let $\sum_{n=1}^{\infty} z_n = L.$ Then for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ for which $n > N$ implies
that $|z_n - L|<\dfrac{\epsilon}{2}$. Thus for $n>N+1$ we have
\begin{equation}
|(z_{n}-L)-(z_{n-1}-L)|\le |z_{n}-L|+|z_{n-1}-L|= \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2}=\epsilon.
\end{equation}
Therefore,
\begin{equation}
\lim_{n \rightarrow \infty} (z_{n}-z_{n-1})=0.
\end{equation}
Now we prove that if the series $\sum_{j=0}^{\infty} c_{j}$ converges, then $c_{j}\rightarrow 0$ as $j \rightarrow \infty$. Let us first define the partial sum of the series as
\begin{equation}
S_{n}:= \sum_{j=0}^{n} c_{j}, \hspace{0.1cm} c_{j}\in \mathbb{C}
\end{equation}
By assumption, the convergence of $\sum_{j=0}^{\infty} c_{j}$ implies that $S_{n}=\sum_{j=0}^{n} c_{j}$ converges for some $n > N$. Thus by (13)
\begin{equation}
\lim_{n \rightarrow \infty} c_{n} = \lim_{n \rightarrow \infty}(S_{n} - S_{n-1})= 0
\end{equation}
We now prove that the series $\sum_{j=0}^{\infty} c^{j}$ diverges if $|c| \ge 1$. Since we want to show that
\begin{equation}
\lim_{n \rightarrow \infty} (S_{n}-S_{n-1})=\lim_{n \rightarrow \infty} c^{j}=0,
\end{equation}
it should follow that $\lim_{n \rightarrow \infty}|c^{j}-0|=|c^{j}|=|c|^{j}=0$. But this is not the case as $|c|\ge 1$.
Thus by the last result, $\sum_{j=0}^{\infty} z^{j}$ diverges on its circle of convergence $|z|=1$. QED
(b)Prove that the power series $\sum_{j=1}^{\infty} \dfrac{z^{j}}{j^{2}}$ converges at every point on its circle of convergence $|z|=1$.
First, we show that $\sum_{j=1}^{\infty} \dfrac{1}{j^{p}}$ converges if $p >1$. Let $p>1$, then for $x$ in the interval $[j-1,j]$, we have
\begin{equation}
x^{p} \le j^{p}.
\end{equation}
From which it follows that
\begin{equation}
\dfrac{1}{j^{p}} \le \int_{j-1}^{j} \dfrac{1}{x^{p}}.
\end{equation}
Result (21) follows from
\begin{equation}
\int_{j}^{j-1} \dfrac{1}{x^{p}} \ge \dfrac{1}{\max x^{p}} = \dfrac{1}{j^{p}}
\end{equation}
Now since we have that
\begin{equation}
\sum_{j=2}^{n} \int_{j-1}^{j} \dfrac{1}{x^{p}} = \int_{1}^{n} \dfrac{1}{x^{p}} = \dfrac{x^{-p+1}}{-p+1}\Biggr|_{1}^{n}= \dfrac{n^{-p+1}-1}{-p+1}= \dfrac{n^{-p+1}}{-p+1}+\dfrac{1}{p-1}
\end{equation}
and $p>1$ implies that $-p<-1 \Longleftrightarrow 1-p<0,$ so that
\begin{equation}
\dfrac{1}{n^{p-1}(-p+1)}+\dfrac{1}{p-1} = \dfrac{1}{p-1}
\end{equation}
as $n \rightarrow \infty.$
Therefore by the comparison test, $\sum_{j=1}^{\infty} \dfrac{1}{j^{p}}$ converges.
Now we prove that the power series $\sum_{j=1}^{\infty} \dfrac{z^{j}}{j^{2}}$ converges at every point on its circle of convergence $|z|=1$.
Since $\dfrac{z^{j}}{j^{2}} = \dfrac{|z^{j}|}{j^{2}}= \dfrac{1}{j^{2}}$ on the circle $|z|=1$, it follows by the last result that $\sum_{j=1}^{\infty} \dfrac{1}{j^{2}}$
converges by the comparison test since $p=2 > 1.$ QED
\stepcounter{enumi}
\item Using the results of Prob. 2, find the circle of convergence of each of the following power series:
\item
\end{enumerate}
\end{document}
给了我以下片段：
我setcounter{enumi}{11}在问题 3 之后进行了尝试，但似乎不起作用。
答案1
\setcounter{enumi}{13}在第 14 项正常工作之前：
\documentclass{article}
\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{graphicx}
\usepackage{setspace}
\usepackage{amsfonts}
\onehalfspacing
\DeclareMathOperator{\Log}{Log}
\DeclareMathOperator{\Arg}{Arg}
\begin{document}
\title{Homework Chapter 5}
\author{}
\maketitle
\setcounter{section}{5}
\setcounter{subsection}{1}
\subsection{Problems 1, 3, and 14}
\begin{enumerate}[wide=0pt, label=\bfseries\arabic*., start=1]
\item (a) Prove that the power series $\sum_{j=0}^{\infty} z^{j}$ converges at no point on its circle of convergence $|z|=1$.
converges by the comparison test since $p=2 > 1.$ QED
\stepcounter{enumi}
\item Using the results of Prob. 2, find the circle of convergence of each of the following power series:
\setcounter{enumi}{13}
\item text
\end{enumerate}
\end{document}
如果所有数字都是这种非线性的，您也可以手动设置它们：\item[\bfseries 14.] text