# strange behavior with \centering

Based on what I had read here, I had expected that the use of \centering would be the right tool for the what I am doing in this code:

\begin{equation*}
\left( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)(n)} = 1 - \frac{1}{n} \right)
\end{equation*}

\\ {\centering implies}

\begin{equation*}
\left( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)(n)} + \frac{1}{(n)(n+1)} = 1 - \frac{1}{n+1} \right);
\end{equation*}
that is, we will prove that if our rule holds good for $n$, then it holds good for $n+1$.


However, when I put the braces about \centering, the result is not centered; when I do, subsequent text is centered as well, which isn't what I want.

How can I get just a single centered line here, evenly spaced between equations?

## 2 Answers

You needn't leave blank lines between the equation* environments, nor use \\; instead you need to terminate the centered paragraph:

\begin{equation*}
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)(n)} = 1 - \frac{1}{n}
\end{equation*}
{\centering implies\par}
\begin{equation*}
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)(n)} + \frac{1}{(n)(n+1)} = 1 - \frac{1}{n+1};
\end{equation*}
that is, we will prove that if our rule holds good for $n$, then it holds good for $n+1$.


However, a much better result can be obtained with the gather* environment provided by amsmath, that will also inhibit page breaks, by default.

\documentclass{article}
\usepackage{amsmath}

\begin{document}
The equation
\begin{gather*}
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)(n)} = 1 - \frac{1}{n}
\\
\text{implies}
\\
\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} \dots + \frac{1}{(n-2) \cdot (n-1)} + \frac{1}{(n-1)n} + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1};
\end{gather*}
that is, we will prove that if our rule holds good for $n$, then it holds good for $n+1$.

\end{document}


You can also say \\[1ex] for increasing the distance between the lines. I've omitted the outer parentheses that don't seem right.

With the standard line width the second equation is overfull, however.

\centering specifies alignment at the paragraph level and in your case } is seen and so al the centering settings are lost before the paragraph ends.

\\\hspace*{\fill}implies\hspace*{\fill}


will centre the text in a line.

Also you should almost always never have a blank line before a display equation as it will generate a blank paragraph just consisting of an indention box and parfillskip glue.