# Complex fractions

How to make this more adequate ('beatifuller' :))? 2 needs to be better shown as being in the power of n-1, but the current representation just smothers it.

\frac{1}{1-\sum_{2}^{n} \frac{1}{{\frac{r}{R}}^(2^(n-1))}}

\documentclass[10pt]{article}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{textcomp}
\begin{document}
\frac{1}{1-\sum_{2}^{n} \frac{1}{{\frac{r}{R}}^(2^(n-1))}}
\end{document}


one more variation:

\documentclass[10pt]{article}
\usepackage{nccmath}

\begin{document}

$\frac{1}{1-\sum\limits_{2}^{n} \Bigl(\mfrac{r}{R}\Bigr)^{\! -2^{n-1}}}$

\end{document}


If you want to write abc, you need to write a^{bc} instead of a^bc or a^(bc). I prefer this way

\documentclass[10pt]{article}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{textcomp}
\begin{document}
$\frac{1}{1-\sum_{2}^{n}\cfrac{1}{\left(\frac{r}{R}\right)^{2^{n-1}}}}$
\end{document}


I propose one of these, based on the medium-size commands from nccmath (~ 80% of \displaystyle):

\documentclass[10pt]{article}
\usepackage{amssymb}
\usepackage{mathtools, nccmath}

\begin{document}

$\frac{1}{1-\medop\sum_{2}^{n} \frac{1}{\bigl(\mfrac{r}{R}\bigr)^{\! 2^{n-1\mathstrut}}}} \qquad \frac{1}{1-\medop\sum_{2}^{n} \bigl(\mfrac{R}{r}\bigr)^{\!2^{n-1}}}$

\end{document}