2
\begin{align*}
\sum_{j=1}^{m_0-1}\,(-1)^{j-1}\hspace{-3ex} 
\sum_{\substack{ k_{m_0 +1} \leq k_{m_0} \leq \cdots \leq k_{j+2} \leq k_1 \leq \boldsymbol{\ell} \\ k_2 < \cdots < k_j < k_{j+1} \leq k_1 }} 
= \sum_{ k_{m_0 +1} \leq k_{m_0} \leq \cdots \leq k_2 \leq k_1 \leq 
 \boldsymbol{\ell}, } \tag(3) \label{eq:3}
\end{align*}
which is very close to our goal, equation \eqref{eq:2}. 
Comparing equation \eqref{eq:3} with equation \eqref{eq:2}, we discover that both have the identical 1\ts{st} through $(m_0 -2)$\ts{th} terms ($1 \leq j \leq m_0-2$). 
The only difference of the two is that the $(m_0 -1)$\ts{th} term in equation \eqref{eq:3} is ...
\end{document}

I try to put a tag next to the equation, but the result comes out as:

enter image description here

The tag is not displayed correctly. Also, the \eqref{eq:3} are displayed in the same, wrong way. (I omit the code having eq:2 and eq:1.)

  • 1
    Welcome to TeX.SE! Replace \tag(3) by \tag{3} (curly brackets around 3). Otherwise LaTeX thinks you mean \tag{(}. – user121799 Dec 30 '18 at 2:57
  • You're welcome! BTW, I'd recommend to move the comma after \boldsymbol{\ell} out of the subscript, and not to abbreviate \textsuperscript by \ts if you ever plan to share your documents with others. – user121799 Dec 30 '18 at 3:11
2

\tag{<tag>} takes a mandatory argument. If you don't supply the surrounding braces, then it takes the next token as the argument.

In your case, \tag(3) doesn't use the suggested format \tag{<tag>}, so it turns into \tag{(}3), numbering the equation with ( and leaving 3) in the input stream. You're looking for \tag{3}:

enter image description here

\documentclass{article}

\usepackage{amsmath,amssymb}

\begin{document}

\begin{align*}
  \sum_{j = 1}^{m_0 - 1}\,(-1)^{j - 1}\hspace{-3ex} 
  \sum_{
    \substack
      { k_{m_0 + 1} \leq k_{m_0} \leq \cdots \leq k_{j + 2} \leq k_1 \leq \boldsymbol{\ell} \\
        k_2 < \cdots < k_j < k_{j+1} \leq k_1 
      }
  } 
  = \sum_{ k_{m_0 + 1} \leq k_{m_0} \leq \cdots \leq k_2 \leq k_1 \leq \boldsymbol{\ell} }, \tag{3} \label{eq:3}
\end{align*}
which is very close to our goal, equation~(2). Comparing equation~\eqref{eq:3} with equation~(2), we discover that both have the identical 
1\textsuperscript{st} through $(m_0 - 2)$\textsuperscript{th} terms ($1 \leq j \leq m_0 - 2$). The only difference of the two is that the 
$(m_0 - 1)$\textsuperscript{th} term in equation \eqref{eq:3} is \ldots

\end{document}

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