如何使 \array 环境中的分数变得很小并且被挤压？

Question

我是一名新 Latex 用户，在 \array 环境中写分数时遇到了问题。文本被压缩到难以阅读我写的内容，到目前为止，我花了几个小时寻找问题的解决方案，但毫无进展。我希望 \array 环境看起来与 \array 环境中没有的方程式相似。

\begin{proof}
    We proceed by induction.
    \begin{enumerate}
        \item Let $n = 1$. Then \[ \frac{1}{1(1+1)} = \frac{1}{2} = \frac{1}{1+1} = \frac{1}{2} \]
    Therefore, the base case holds.
        \item Assume now by Induction that \[ \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4} + \dotsb + \frac{1}{n(n+1)}=\frac{n}{n+1} \] holds for some $n \in{\mathbb{N}}$. We now show that \[ \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4} + \dotsb + \frac{1}{(n+1)(n+2)}=\frac{n+1}{n+2} \] We write \[ \begin{array}{lll}
            \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3} + \dotsb + \frac{1}{(n+1)(n+2)} & = & \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3} + \dotsb + \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)}  \\ 
             & = & \frac{n}{n+1} + \frac{1}{(n+1)(n+2)} \\
             & = & \frac{(n+1)^{2}}{(n+1)(n+2)} \\
             & = & \frac{n + 1}{n + 2}.
        \end{array} \]
        \end{enumerate}
By Induction, \[ \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4} + \dotsb + \frac{1}{(n+1)(n+2)}=\frac{n+1}{n+2} \] for every $n \in{\mathbb{N}}$.  

Answer 1

您应该使用align*而不是array：

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

\begin{document}

\begin{proof}
We proceed by induction.
\begin{enumerate}
\item Let $n = 1$. Then 
\[
\frac{1}{1(1+1)} = \frac{1}{2} = \frac{1}{1+1} = \frac{1}{2}
\]
Therefore, the base case holds.

\item Assume now by induction that 
\[
\frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4} 
+ \dots + \frac{1}{n(n+1)}=\frac{n}{n+1}
\]
holds for some $n \in \mathbb{N}$. We now show that
\[
\frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}
+ \dots + \frac{1}{(n+1)(n+2)}=\frac{n+1}{n+2}
\]
We write
\begin{align*}
\frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}
  + \dots + \frac{1}{(n+1)(n+2)}
& = \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}
    + \dots + \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} \\ 
& = \frac{n}{n+1} + \frac{1}{(n+1)(n+2)} \\
& = \frac{(n+1)^{2}}{(n+1)(n+2)} \\
& = \frac{n + 1}{n + 2}.
\end{align*}
\end{enumerate}
By induction,
\[
\frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}
+ \dots + \frac{1}{(n+1)(n+2)}=\frac{n+1}{n+2}
\]
for every $n \in \mathbb{N}$.  
\end{proof}

\clearpage

\begin{proof}
We proceed by induction. Set
\[
S(n)=\frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4} 
     + \dots + \frac{1}{n(n+1)}
\]
for simplicity.

\begin{enumerate}
\item Let $n = 1$. Then 
\[
S(1) = \frac{1}{1(1+1)} = \frac{1}{2} = \frac{1}{1+1}
\]
Therefore, the base case holds.

\item Assume now by induction that 
\[
S(n) = \frac{n}{n+1}
\]
holds for some $n \in \mathbb{N}$. We now show that
\[
S(n+1) = \frac{n+1}{n+2}
\]
We write
\begin{align*}
S(n+1)
& = S(n) + \frac{1}{(n+1)(n+2)} \\
& = \frac{n}{n+1} + \frac{1}{(n+1)(n+2)} \\
& = \frac{(n+1)^{2}}{(n+1)(n+2)} \\
& = \frac{n+1}{n+2}.
\end{align*}
\end{enumerate}
By induction,
\[
S(n) = \frac{1}{1 \cdot 2}+ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}
+ \dots + \frac{1}{(n+1)(n+2)}=\frac{n+1}{n+2}
\]
for every $n \in \mathbb{N}$.  
\end{proof}


\end{document}

当然，长队是一个大问题，所以我建议采用替代版本。