# Can \left( … \right) be instructed to ignore nonexistent upper or lower limits of summation? [duplicate]

In a formula such as

$\left( \sum_{n\in\mathbf{N}} \right)$


the left and right delimiters extend not only downwards to cover the lower limit of the sum, but upwards an equal amount as if to cover an upper limit, despite there being no upper limit. Is there a way to prevent this?

This seems related to this question, and the first answer there (enclose it in an array environment) can sort of solve it, but (I think) inelegantly, and also inserts extra spacing. I'm hoping there is a better way, like a command to simply control whether \sum has a top delimiter at all or something.

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## marked as duplicate by Mike Shulman, Herbert, mafp, Thorsten, zerothApr 16 '13 at 18:44

When dealing with the issue of placing large parentheses around a displaystyle-sized summation symbol in one of the math-related chapters of the TeXbook, Knuth explicitly recommends using \biggl( ... \biggr) rather than \left( ... \right). I.e., the size of the "fences" chosen by \left( ... \right) are too large. – Mico Apr 16 '13 at 13:51
I agree, generally size the fences such that it is clear what they fence in, but not such that they dominate the expression. As for sums, there is no need for the fences to encapsulate the limits. – daleif Apr 16 '13 at 13:54
The issue isn't so much that the parentheses are too large, it's that they are off-center vertically: they include lots of empty space at the top. Manually choosing a size doesn't change that. – Mike Shulman Apr 16 '13 at 14:33
@MikeShulman - I'd argue that TeX is not placing the parentheses off-center vertically, irrespective of their (insufficient, correct, or exaggerated) size: note that the "bellies" of the parentheses are lined up with the middle of the large operator symbol in all cases. Moreover, I'd argue that it's neither necessary nor desirable to enclose the (lower and/or upper) limits of summation with parentheses, lest the fences dominate the entire expression. Hence the recommendation to use \bigg as the explicit size instruction. – Mico Apr 16 '13 at 14:44
I kind of object to being told that my question is wrong. This is what I want to do -- the question is whether it can be done. Personally, I think it looks ugly and confusing not to enclose the limits of summation in the parentheses. – Mike Shulman Apr 16 '13 at 15:50

Is it one of the following?

\documentclass{article}

\usepackage{amsmath}

\begin{document}

$$\Bigl(\sum_{n\in\mathbf{N}}\Bigr) \biggl(\sum_{n\in\mathbf{N}}\biggr)$$

\end{document}


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Since the fence size produced by \Biggl( ... \Biggr) is the same as that produced by \left( ... \right), and given that the OP has already declared that size to be too large, the third option should definitely not be considered. – Mico Apr 16 '13 at 14:21
@Mico Good point; I'll edit my answer. – Svend Tveskæg Apr 16 '13 at 14:26
The issue isn't just that the parentheses are too large, it's that they are off-center vertically. – Mike Shulman Apr 16 '13 at 14:32

Parentheses are placed symmetrically with respect to the math axis, an imaginary line a bit over the baseline, that can be seen by typesetting a fraction, because the fraction line lies exactly on the math axis.

A problem with fences around an expressions such as

\sum_{n\in\mathbf{N}} 2^n
\sum_{n\in\mathbf{N}} \frac{2^n-1}{2^n+1}


depends on the whole expression to be enclosed.

After observing that \left( and \right) will give poor results, the choice can be between \Big and \bigg. Here's a visual comparison.

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{gather*}
\Bigl(\,\sum_{n\in\mathbf{N}} 2^n\Bigr)
\biggl(\,\sum_{n\in\mathbf{N}} 2^n\biggr)
\$6pt] \Bigl(\,\sum_{n\in\mathbf{N}} \frac{2^n-1}{2^n+1}\Bigr) \quad \biggl(\,\sum_{n\in\mathbf{N}} \frac{2^n-1}{2^n+1}\biggr) \end{gather*} \end{document}  I'd probably use \Big for the first kind of formula and \bigg for the second one. Something like \[ 2+\left( \raisebox{1.5mm}{\displaystyle\sum_{n\in\mathbf{N}}2^{-n}} \right)=4$


that produces

is clearly unacceptable. Nor is better

$2+\raisebox{-1.5mm}{\displaystyle \left( \raisebox{1.5mm}{\displaystyle\sum_{n\in\mathbf{N}}2^{-n}} \right)}=4$


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As I said, what I want is for the delimiters to enclose the lower limit, but not any extra space for the upper limit. – Mike Shulman Apr 16 '13 at 16:17
@MikeShulman I find no reason for this. – egreg Apr 16 '13 at 16:34
I'm sorry to hear that. – Mike Shulman Apr 16 '13 at 17:12
@MikeShulman Look at the new examples to see what I mean. A parenthesized subformula makes sense only when set in context. – egreg Apr 16 '13 at 17:53
@Mike (and egreg, but you'll be notified automatically): It seems to me that the last option might well look better with double vertical bar delimiters rather than parentheses. The OP has said that in the case he really cares about, he's using double bar delimiters. – Charles Staats Apr 16 '13 at 20:11

The following seems to do what I want in the test case, at least:

$\left( \raisebox{1.5mm}{\displaystyle\sum_{n\in\mathbf{N}}} \right)$


But I don't understand why, and it seems a bit kludgy.

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I wouldn't recommend such a construction, which is more kludgy than using \biggl and \biggr. – egreg Apr 16 '13 at 15:34
Since \biggl and \biggr don't solve the problem, their relative kludginess seems irrelevant. – Mike Shulman Apr 16 '13 at 15:50
You surely have something beside the summation sign. – egreg Apr 16 '13 at 16:01
Yeah, so it has to go in the \raisebox` too, I guess. – Mike Shulman Apr 16 '13 at 16:10