在对齐环境中垂直对齐 \vdots - 将 \vdots 偏移“+”宽度的一半

Question

在align环境中，我有任意数量的项的总和。我想使用\vdots显示省略号，表示存在未显示的项。由于我垂直对齐+，因此我希望显示的省略号\vdots相对于 居中+。在 中TikZ，有命令（如\newlength\width_of_plus_sign和 ）\settowidth{\width_of_plus_sign}{$+$}可以执行此操作。有与\align环境类似的环境吗？

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

\begin{document}


For each integer $2 \leq i \leq n$,
\begin{align*}
\frac{1}{n^{i}} \binom{n}{i} &= \frac{1}{n^{i}} \frac{n!}{i!(n-i)!} \\
&= \frac{1}{n^{i}} \frac{n(n - 1)(n - 2) \cdots (n - i + 1)}{i!} \\
&= \frac{1}{i!} \frac{n(n - 1) (n - 2) \cdots (n - (i - 1))}{n^{i}} \\
&= \frac{1}{i!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) .
\end{align*}
So,
\begin{align*}
&\sum_{i=k+1}^{n} \frac{1}{n^{i}} \binom{n}{i} \\
&\qquad = \sum_{i=k+1}^{n} \frac{1}{i!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) \\
&\qquad = \frac{1}{(k+1)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k}{n}\right) \\
&\qquad\qquad \!\begin{aligned}[t]
&+ \frac{1}{(k+2)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k+1}{n}\right) \\
&+ \frac{1}{(k+3)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k+2}{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) .
\end{aligned} \\
&\qquad = \frac{1}{k!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k}{n}\right) \\
&\qquad\qquad \!\begin{aligned}[t]
&\left[\frac{1}{k+1} \left(1 - \frac{k}{n}\right) \right. \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}+ \frac{1}{(k+1)(k+2)}  \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}+ \frac{1}{(k+1)(k+2)(k+3)}  \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \left(1 - \frac{k+2}{n}\right) \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}+ \ldots \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}\left.
+ \frac{1}{(k+1)(k+2) \cdots n} \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \cdots \left(1 - \frac{n-1}{n}\right)
\right] .
\end{aligned}

\end{document}

Answer 1

通过一些简化，既可以简化输入，也可以获得预期的输出：

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

\newcommand{\pterm}[1]{\left(1-\frac{#1}{n}\right)}
\newcommand{\inv}[1]{\frac{1}{#1}}

\begin{document}

\begin{align*}
\mathmakebox[2em][l]{\sum_{i=k+1}^{n} \inv{n^{i}} \binom{n}{i} \lvert x \rvert^{i}} \\
&= \sum_{i=k+1}^{n} \inv{i!} \pterm{1} \pterm{2} \dots \pterm{i - 1} \\
&= \inv{(k+1)!} \pterm{1} \pterm{2} \dots \pterm{k} \\
&\qquad\!\begin{aligned}[t]
  &+ \inv{(k+2)!} \pterm{1} \pterm{2} \dots \pterm{k+1} \\
  &+ \inv{(k+3)!} \pterm{1} \pterm{2} \dots \pterm{k+2} \\
  &+ \dots + \inv{n!} \pterm{1} \pterm{2} \dots \pterm{n-1} .
\end{aligned} \\
&= \!\begin{aligned}[t]
  \inv{k!}
    & \pterm{1} \pterm{2} \dots \pterm{k} \\
    &\biggl[\inv{k+1} \pterm{k}\\
    &\hphantom{\biggl[} + \inv{(k+1)(k+2)}  \pterm{k} \pterm{k+1} \\
    &\hphantom{\biggl[}+ \inv{(k+1)(k+2)(k+3)}  \pterm{k} \pterm{k+1} \pterm{k+2} \\
    &\phantom{\biggl[}+ {} \\[-2ex]
    &\hphantom{\biggl[}\vdotswithin{+} \\[-2ex]
    &\phantom{\biggl[}+ {}\\
    &\hphantom{\biggl[}
     + \inv{(k+1)(k+2) \dots n} \pterm{k} \pterm{k+1} \dots \pterm{n-1}\biggr] .
\end{aligned}
\end{align*}

\end{document}

我把所有\ldots和\cdots都改为\dots；你似乎以与平常不同的方式使用它们。

Answer 2

以下代码是受 的评论启发而编写的daleif。我不得不将其添加mathtools到序言中。

\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathtools}

\begin{document}

\begin{align*}
&\sum_{i=k+1}^{n} \frac{1}{n^{i}} \binom{n}{i} \vert x \vert^{i} \\
&\qquad = \sum_{i=k+1}^{n} \frac{1}{i!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) \\
&\qquad = \frac{1}{(k+1)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k}{n}\right) \\
&\qquad\qquad \!\begin{aligned}[t]
&+ \frac{1}{(k+2)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k+1}{n}\right) \\
&+ \frac{1}{(k+3)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k+2}{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) .
\end{aligned} \\
&\qquad = \frac{1}{k!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k}{n}\right) \\
&\qquad\qquad \!\begin{aligned}[t]
&\left[\frac{1}{k+1} \left(1 - \frac{k}{n}\right) \right. \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}+ \frac{1}{(k+1)(k+2)}  \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}+ \frac{1}{(k+1)(k+2)(k+3)}  \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \left(1 - \frac{k+2}{n}\right) \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}+ \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}\vdotswithin{+} \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}+ \\
&\hphantom{\left[\vphantom{\frac{1}{k+1}}\right.}\left.
+ \frac{1}{(k+1)(k+2) \cdots n} \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \cdots \left(1 - \frac{n-1}{n}\right)
\right] .
\end{aligned}
\end{align*}

\end{document}