[Tex/LaTex] Insert left-aligned text in split environment (mathtools isn’t allowed)

Question

As I say, I can't use mathtools package or anything else (because my teacher at the university doesn't allow it). I'm trying to write a text (a word, 'és', what means 'and', but it is irrelevant) to the left side. If I just write with a \text command, like this:

\begin{equation*}
  \begin{split}
    D\left(n\right) &= \sum_{m=1}^{n} \left[\frac{n}{m}\right] \geq \sum_{m=1}^{n} \left( \frac{n}{m} - 1 \right) = - n + n \sum_{m=1}^{n} \frac{1}{m} > \\
    &> n \log{\left(n+1\right)} - n > n \log n - n \\
    \text{és} \\
    D\left(n\right) &= \sum_{m=1}^{n} \left[ \frac{n}{n} \right] \leq\sum_{m=1}^n \frac{1}{m} < n\left(1 + \log n \right).
  \end{split}
\end{equation*}

and the output is definitely not on the left:

With amsmath's intertext command:

\begin{equation*}
  \begin{split}
    D\left(n\right) &= \sum_{m=1}^{n} \left[\frac{n}{m}\right] \geq \sum_{m=1}^{n} \left( \frac{n}{m} - 1 \right) = - n + n \sum_{m=1}^{n} \frac{1}{m} > \\
    &> n \log{\left(n+1\right)} - n > n \log n - n \\
    \intertext(és)
    D\left(n\right) &= \sum_{m=1}^{n} \left[ \frac{n}{n} \right] \leq\sum_{m=1}^n \frac{1}{m} < n\left(1 + \log n \right).
  \end{split}
\end{equation*}

the result is another wrong output, because everything is on the left:

So, how to write a word to the left side without additional packages if two separate environment isn't an option because the equations are aligned to each other?

Answer 1

instead of equation*/split try to use align*. And, don't overuse \left and \right to auto-size parentheses.

\documentclass[12pt]{article}
\usepackage{amsmath}

%-------------------------------------- only for show page layout
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.25pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%

\begin{document}

\begin{align*}
D(n)
    & = \sum_{m=1}^{n} \left[\frac{n}{m}\right] \geq \sum_{m=1}^{n} \left( \frac{n}{m} - 1 \right)
        = - n + n \sum_{m=1}^{n} \frac{1}{m} > \\
    & > n \log(n+1) - n > n \log n - n
\intertext{es}
    &= \sum_{m=1}^{n} \left[ \frac{n}{n} \right] \leq\sum_{m=1}^n \frac{1}{m} < n(1 + \log n).
\end{align*}
\end{document}