Can't identify the problem with my code, please help me!

Question

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}

\renewcommand{\qedsymbol}{$\blacksquare$}
\newcommand*{\QEDA}{\hfill\ensuremath{\blacksquare}}%

\begin{document}

\begin{proof}
  Since $x_k$ is only conditionally summable, this means that the
  sequence $s_n = \sum\limits_{k=1}^{n}|x_k|$ is not convergent. Note
  that $s_n$ is a monotone increasing sequence that is not convergent,
  hence it must be unbounded. But
  $\sigma_n^+ = \sum\limits_{k=1}^{n}x_k^+$ and
  $\sigma_n^- = \sum\limits_{k=1}^{n}x_^-$ are all subsequences of
  $s_n$, and they are monotone increasing as well. Thus, they must me
  also unbounded and
  $\sum\limits_{k=1}^{\infty}x_k^+ = \infty =
  \sum\limits_{k=1}^{\infty}x_k^-$.
\end{proof}

\end{document}

I get the following error when I try to run my code, but I can't really find anything wrong with my code.

! Missing { inserted.
<to be read again> 
                   ^
l.19   $\sigma_n^- = \sum\limits_{k=1}^{n}x_^
                                             -$ are all subsequences of
?

Answer 1

It's worth noting that in this case, the error message tells you where the problem is in your code

! Missing { inserted.
<to be read again> 
                   ^
l.19   $\sigma_n^- = \sum\limits_{k=1}^{n}x_^
                                             -$ are all subsequences of
?

The line is broken at the point where LaTeX ran into problems.

x_^- is your problem, although I think your troubleshooting is perhaps not helped by not putting braces around your subscripts and superscripts. This is an issue that comes up from time-to-time on here and I often recommend that you always use braces, even in the case of single-character sub-/superscripts, i.e.: x_{k}. This is a bit more of a pain to type it is true, and it could be argued that the braces are unnecessary clutter. On the other hand, it does make things easier for me to parse visually and you will never have the problem of forgetting to put braces around the superscript when you realise that x^n should have been x^{n + 1}. It's a good habit to get into and I think x_{k}^{+} is more immediately meaningful than x_k^+. To each their own, but just something to think about.

Anyway, if we apply the correction Christian Hupfer suggests in the comments:

$\sigma_n^- = \sum\limits_{k=1}^{n}x_k^-$ are all subsequences of

As in:

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}

\renewcommand{\qedsymbol}{$\blacksquare$}
\newcommand*{\QEDA}{\hfill\ensuremath{\blacksquare}}%

\begin{document}

\begin{proof}
  Since $x_k$ is only conditionally summable, this means that the
  sequence $s_n = \sum\limits_{k=1}^{n}|x_k|$ is not convergent. Note
  that $s_n$ is a monotone increasing sequence that is not convergent,
  hence it must be unbounded. But
  $\sigma_n^+ = \sum\limits_{k=1}^{n}x_k^+$ and
  %% problem line below %%
  $\sigma_n^- = \sum\limits_{k=1}^{n}x_k^-$ are all subsequences of
  %% problem line above %%
  $s_n$, and they are monotone increasing as well. Thus, they must me
  also unbounded and
  $\sum\limits_{k=1}^{\infty}x_k^+ = \infty =
  \sum\limits_{k=1}^{\infty}x_k^-$.
\end{proof}

\end{document}

All is well. Although I would urge you to move away from your pervasive use of \limits. In-line math mode sets the sub- and superscripts to the side for a very good reason, and that is preserving consistent line-spacing, with the appropriate amount of white space between the lines.