# How to make Fractions in \array environment from being tiny and squishing up? [duplicate]

I am a new Latex user who's having problems when writing fractions in the \array environment. The text becomes compressed to where reading what I wrote becomes a challenge, and I have so far spent hours finding a solution to my problem with no luck. I want to the \array environment to look similar to the equations not in the \array environment.

\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 $n \in{\mathbb{N}}$.

• You could use \dfrac (short hand for \displaystyle\frac) – user94293 Jul 24 '18 at 6:27
• Related to tex.stackexchange.com/questions/32824/… It's because the math style changes from display to text mode. – nox Jul 24 '18 at 6:27
• Replace \begin{array}{lll} with \begin{align*}, \end{array} with \end{align*}; moreover & = & should become &=. – egreg Jul 24 '18 at 7:25
• your main error is using \begin{array} which is designed for matrices/arrays of values not for setting a displayed equation. – David Carlisle Jul 24 '18 at 7:48
• @user94293 actually if you find yourself needing \dfrac that is usually a sign of an error elsewhere (in this case using array instead of align*) – David Carlisle Jul 24 '18 at 7:49

## 1 Answer

You should use align* instead of 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}


Of course, the long lines are a big problem, so I suggest an alternative version. 