6

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?

9

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.

9

\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.

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