That's expected, because \mathclap makes a zero width box, so TeX cannot “see” the width of those limits.
I wouldn't use \left and \right, which produce too big delimiters in this case, but \biggl and \biggr, with a little adjustment.
Here I present a few realizations.
- (1) is clearly wrong;
- (2) is also wrong because of the too large size and the unwanted space between the fraction and the parenthesis;
- (3) is better as regards to size and spacing, but the parenthesis clashes with the subscript;
- (4) fixes the clash;
- (5) fixes the excess space between the summation and the terms to sum;
- (6) does the same as (5), but the backing up is too much, in my opinion.
Take your pick. Remember that automation and fine typography don't go along well.
\documentclass{article} \usepackage{mathtools} \begin{document} \begin{alignat}{2} S_1 &= \frac{1}{2} \left(\sum_{\mathclap{i = n + 1}}^{n + m} a_i t_i \right) &\qquad& \text{your attempt} \\ S_1 &= \frac{1}{2} \left(\sum_{i = n + 1}^{n + m} a_i t_i \right) &\qquad& \text{without \texttt{\string\mathclap}} \\ S_1 &= \frac{1}{2} \biggl(\sum_{i = n + 1}^{n + m} a_i t_i \biggr) && \text{with \texttt{\string\bigg}} \\ S_1 &= \frac{1}{2} \biggl(\,\sum_{i = n + 1}^{n + m} a_i t_i \biggr) && \text{with the needed correction} \\ S_1 &= \frac{1}{2} \biggl(\,\sum_{i = n + 1}^{n + m} \hspace{-0.33em} a_i t_i \biggr) && \text{with some back space} \\ S_1 &= \frac{1}{2} \biggl(\,\smashoperator[r]{\sum_{i = n + 1}^{n + m}} a_i t_i \biggr) && \text{with \texttt{\string\smashoperator}} \end{alignat} \end{document}

\mathclapif you don't want that.\leftmakes an over-sized ( here if you used\bigl(the ( wouldn't be interfering with the subscript