在 alignat 环境中未获得我想要的对齐

Question

在以下显示中，环境$+ \frac{1}{4!}$的第三行alignat应在第二行的正下方对齐$+ \frac{1}{2!}$。根据更正后的显示中的间距，我可能希望将其$+ \frac{1}{n!}$直接对齐到正下方$+ \frac{1}{4!}$。为什么我的代码没有提供这种对齐方式？我如何获得我想要的对齐方式？

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amsfonts}

\begin{document}

\noindent {\em The sequence}
\begin{equation*}
\left(1 + \frac{1}{n}\right)^{n}
\end{equation*}
{\em is bounded between $2$ and $2.75$.}
\vskip0.2in

\noindent $a_{1} = 2$, and since the given sequence is an increasing sequence, $a_{n} > 2$ for every integer $n > 1$. According to the Binomial Theorem,
\allowdisplaybreaks{
\begin{alignat*}{2}
\left(1 + \frac{1}{n}\right)^{n} &= 1 + \binom{n}{1} \frac{1}{n} + \binom{n}{2} \frac{1}{n^{2}} + \binom{n}{3} \frac{1}{n^{3}} + \ldots
+ \binom{n}{n} \frac{1}{n^{n}} \\
&= 1 + 1 &&+ \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \\
&&&+ \frac{1}{4!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\left(1 - \frac{3}{n}\right) + \ldots \\
&&&+ \frac{1}{n!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{n - 1}{n}\right) \\
&< 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} \\
&= 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{2}{3!} + \frac{2}{4!} + \ldots + \frac{2}{n!}\right) \\
&< 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{1}{3} + \frac{1}{3^{2}} + \ldots + \frac{1}{3^{n-2}}\right) \\
&< 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{1}{3} + \frac{1}{3^{2}} + \ldots + \frac{1}{3^{n-2}} + \frac{1}{3^{n-1}} + \frac{1}{3^{n}} + \ldots \right) \\
&< 1 + 1 + \frac{1}{2} + \frac{1}{2} \left( \frac{1}{2} \right) \\
&= 2 + \frac{3}{4} . \ \rule{1.5ex}{1.5ex}
\end{alignat*}}

\end{document}

Answer 1

我认为没有必要设置和微调alignat*环境。一个基本的align*环境，加上两个\phantom语句就足够了。

\documentclass{amsart}
\begin{document}
\noindent 
{\em The sequence
\begin{equation*}
\left(1 + \frac{1}{n}\right)^{n}
\end{equation*}
is bounded between $2$ and $2.75$.}

\vskip0.2in

\noindent $a_{1} = 2$, and since the given sequence is an increasing 
sequence, $a_{n} > 2$ for every integer $n > 1$. According to the Binomial 
Theorem,
\allowdisplaybreaks
\begin{align*}
\left(1 + \frac{1}{n}\right)^{n} &= 1 + \binom{n}{1} \frac{1}{n} + 
  \binom{n}{2} \frac{1}{n^{2}} + \binom{n}{3} \frac{1}{n^{3}} + \dots
  + \binom{n}{n} \frac{1}{n^{n}} \\
&= 1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - 
  \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \\
&\phantom{{}= 1 + 1}{}+ \frac{1}{4!}\left(1 - \frac{1}{n}\right)\left(1 - 
  \frac{2}{n}\right)\left(1 - \frac{3}{n}\right) + \dotsb \\
&\phantom{{}= 1 + 1}{}+ \frac{1}{n!}\left(1 - \frac{1}{n}\right)\left(1 - 
  \frac{2}{n}\right) \dotsb \left(1 - \frac{n - 1}{n}\right) \\
&< 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!} \\
&= 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{2}{3!} + \frac{2}{4!} + 
  \dots + \frac{2}{n!}\right) \\
&< 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{1}{3} + \frac{1}{3^{2}} + 
  \dots + \frac{1}{3^{n-2}}\right) \\
&< 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{1}{3} + \frac{1}{3^{2}} + 
\dots + \frac{1}{3^{n-2}} + \frac{1}{3^{n-1}} + \frac{1}{3^{n}} + \dotsb 
  \right) \\
&< 1 + 1 + \frac{1}{2} + \frac{1}{2} \left( \frac{1}{2} \right) \\
&= 2 + \frac{3}{4} \,. \quad \rule{1.5ex}{1.5ex}
\end{align*}

\end{document}

Answer 2

嵌套aligned环境align：

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{amssymb}

\newtheorem*{statement}{Statement}

\allowdisplaybreaks

\begin{document}

\begin{statement}
The sequence
\begin{equation*}
a_{n}=\left(1 + \frac{1}{n}\right)^{n}
\end{equation*}
is bounded between $2$ and $2.75$.
\end{statement}

\begin{proof}
We have $a_{1} = 2$ and, since the given sequence is an increasing sequence, $a_{n} > 2$ 
for every integer $n > 1$. According to the Binomial Theorem,
\begin{align*}
\left(1 + \frac{1}{n}\right)^{n} 
  &= 1 + \binom{n}{1} \frac{1}{n} + \binom{n}{2} \frac{1}{n^{2}} + \binom{n}{3} \frac{1}{n^{3}} + \dots
     + \binom{n}{n} \frac{1}{n^{n}} \\
  &= \!\begin{aligned}[t]
     1 + 1 &+ \frac{1}{2!}\left(1 - \frac{1}{n}\right)
            + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \\
           &+ \frac{1}{4!}\left(1 - \frac{1}{n}\right)\left(1 
            - \frac{2}{n}\right)\left(1 - \frac{3}{n}\right) + \dotsb \\
           &+ \frac{1}{n!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \dots
                \left(1 - \frac{n - 1}{n}\right)
     \end{aligned}\\
  &< 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots + \frac{1}{n!} \\
  &= 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{2}{3!} + \frac{2}{4!} + \dots + \frac{2}{n!}\right) \\
  &< 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{1}{3} + \frac{1}{3^{2}} + \dots
       + \frac{1}{3^{n-2}}\right) \\
  &< 1 + 1 + \frac{1}{2!} + \frac{1}{2}\left(\frac{1}{3} + \frac{1}{3^{2}} + \ldots + \frac{1}{3^{n-2}}
       + \frac{1}{3^{n-1}} + \frac{1}{3^{n}} + \ldots \right) \\
  &< 1 + 1 + \frac{1}{2} + \frac{1}{2} \left( \frac{1}{2} \right) \\
  &= 2 + \frac{3}{4} .\qedhere
\end{align*}
\end{proof}

\end{document}

请注意，这\allowdisplaybreaks是声明，而不是带参数的命令。它应该在序言中发布，并在文档的最终版本中删除，并\displaybreak在所需的位置添加合适的命令。