如何加宽页面

Question

\documentclass{article}
\usepackage{amsmath, amssymb}
\usepackage[many]{tcolorbox}
\usepackage{lipsum}
\DeclareMathOperator{\dint}{\displaystyle\int}
\DeclareMathOperator{\dsum}{\displaystyle\sum}
\definecolor{myblue}{RGB}{0,163,243}

\newtcolorbox[auto counter,number within=section]{exo}[1][]{
  enhanced jigsaw,colback=white,colframe=myblue,coltitle=myblue,
  fonttitle=\bfseries\sffamily,
  sharp corners,
  detach title,
  leftrule=18mm,
  underlay unbroken and first={\node[below,text=white,font=\sffamily\bfseries,align=center]
    at ([xshift=-11mm,yshift=-1mm]interior.north west) {Exercice\\\thetcbcounter};},
  breakable,pad at break=1mm,
  #1,
  code={\ifdefempty{\tcbtitletext}{}{\tcbset{before upper={\tcbtitle\par\medskip}}}},
}


\begin{document}

\begin{exo}
\begin{enumerate}
\item Soit $\alpha >0$ , Montrer que $\dint_{0}^{1}\dfrac{dt}{1+t^{\alpha }}=%
\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{1+n\alpha }$ .
En d\'{e}duire les sommes 
\begin{equation*}
\overset{+\infty }{\underset{n\,=1}{\sum }}\dfrac{(-1)^{n}}{n}~;\overset{%
+\infty }{\underset{n\,=\,0}{\sum }}\dfrac{(-1)^{n}}{1+2n}\, \ 
\end{equation*}%
Indication : On pourra utiliser que\textit{\ }$\dfrac{1}{1+u}=\overset{n}{%
\underset{k\,=\,0}{\dsum }}\left( -1\right) ^{k}u^{k}+\dfrac{\left(
-1\right) ^{n+1}u^{n+1}}{1+u}$

\item Calculer $\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{%
1+n}$ et $\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{1+2n}$

\item On pose , pour $\alpha >0$ , $R_{n}\left( \alpha \right) =\overset{%
+\infty }{\underset{k=n+1}{\dsum }}\dfrac{(-1)^{k}}{1+k\alpha }$ , montrer
que :%
\begin{equation*}
R_{n}\left( \alpha \right) =\left( -1\right) ^{n+1}\int_{0}^{1}\dfrac{%
t^{\left( n+1\right) \alpha }}{1+t^{\alpha }}dt=\dfrac{\left( -1\right)
^{n+1}}{2\left( n+1\right) \alpha +2}+\dfrac{\alpha \left( -1\right) ^{n+1}}{%
\left( n+1\right) \alpha +1}\int_{0}^{1}\dfrac{t^{\left( n+2\right) \alpha }%
}{\left( 1+t^{\alpha }\right) ^{2}}dt
\end{equation*}%
Montrer que $R_{n}\left( \alpha \right) \underset{+\infty }{\sim }\dfrac{%
\left( -1\right) ^{n+1}}{2\left( n+1\right) \alpha +2}.$

\item Etudier la nature de la s\'{e}rie $\sum R_{n}\left( \alpha \right) .$
\end{enumerate}
\end{exo}
\end{document}

Answer 1

例如，您可以使用环境split来将长方程除以=符号，如下（参见标有 的添加代码<======）：

\begin{equation*}
\begin{split} % <=======================================================
R_{n}\left( \alpha \right) &=\left( -1\right) ^{n+1}\int_{0}^{1}\dfrac{% <========================
t^{\left( n+1\right) \alpha }}{1+t^{\alpha }}dt \\ % <==================
&=\dfrac{\left( -1\right) % <===========================================
^{n+1}}{2\left( n+1\right) \alpha +2}+\dfrac{\alpha \left( -1\right) ^{n+1}}{%
\left( n+1\right) \alpha +1}\int_{0}^{1}\dfrac{t^{\left( n+2\right) \alpha }%
}{\left( 1+t^{\alpha }\right) ^{2}}dt
\end{split} % <=========================================================
\end{equation*}%

您的代码中有几个错误我忽略了，因为我们不知道您是如何定义相关命令的\dint。如果您将定义添加到您的问题中，我可以更新我的答案。

请参阅以下 MWE

\documentclass{article}

\usepackage{amsmath, amssymb}
\usepackage[many]{tcolorbox}
\usepackage{lipsum}

\definecolor{myblue}{RGB}{0,163,243}

\newtcolorbox[auto counter,number within=section]{exo}[1][]{
  enhanced jigsaw,colback=white,colframe=myblue,coltitle=myblue,
  fonttitle=\bfseries\sffamily,
  sharp corners,
  detach title,
  leftrule=18mm,
  underlay unbroken and first={\node[below,text=white,font=\sffamily\bfseries,align=center]
    at ([xshift=-11mm,yshift=-1mm]interior.north west) {Exercice\\\thetcbcounter};},
  breakable,pad at break=1mm,
  #1,
  code={\ifdefempty{\tcbtitletext}{}{\tcbset{before upper={\tcbtitle\par\medskip}}}},
}


\begin{document}

\begin{exo}
\begin{enumerate}
\item Soit $\alpha >0$ , Montrer que $\dint_{0}^{1}\dfrac{dt}{1+t^{\alpha }}=%
\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{1+n\alpha }$ .
En d\'{e}duire les sommes 
\begin{equation*}
\overset{+\infty }{\underset{n\,=1}{\sum }}\dfrac{(-1)^{n}}{n}~;\overset{%
+\infty }{\underset{n\,=\,0}{\sum }}\dfrac{(-1)^{n}}{1+2n}\, \ 
\end{equation*}%
Indication : On pourra utiliser que\textit{\ }$\dfrac{1}{1+u}=\overset{n}{%
\underset{k\,=\,0}{\dsum }}\left( -1\right) ^{k}u^{k}+\dfrac{\left(
-1\right) ^{n+1}u^{n+1}}{1+u}$

\item Calculer $\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{%
1+n}$ et $\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{1+2n}$

\item On pose , pour $\alpha >0$ , $R_{n}\left( \alpha \right) =\overset{%
+\infty }{\underset{k=n+1}{\dsum }}\dfrac{(-1)^{k}}{1+k\alpha }$ , montrer
que :%
\begin{equation*}
\begin{split} % <=======================================================
R_{n}\left( \alpha \right) &=\left( -1\right) ^{n+1}\int_{0}^{1}\dfrac{% <========================
t^{\left( n+1\right) \alpha }}{1+t^{\alpha }}dt \\ % <==================
&=\dfrac{\left( -1\right) % <===========================================
^{n+1}}{2\left( n+1\right) \alpha +2}+\dfrac{\alpha \left( -1\right) ^{n+1}}{%
\left( n+1\right) \alpha +1}\int_{0}^{1}\dfrac{t^{\left( n+2\right) \alpha }%
}{\left( 1+t^{\alpha }\right) ^{2}}dt
\end{split} % <=========================================================
\end{equation*}%
Montrer que $R_{n}\left( \alpha \right) \underset{+\infty }{\sim }\dfrac{%
\left( -1\right) ^{n+1}}{2\left( n+1\right) \alpha +2}.$

\item Etudier la nature de la s\'{e}rie $\sum R_{n}\left( \alpha \right) .$
\end{enumerate}
\end{exo}
\end{document}

及其结果：

Answer 2

我建议将一个方程拆分，并删除许多不必要的\left \right对。无关：我不明白为什么你不直接在键盘上输入重音字母，尤其是因为所有现代 TeX 编辑器都理解 utf8。我使用enumitem、mathtools和进行了一些改进nccmath。

\documentclass[french]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{babel}
\usepackage{mathtools, nccmath, amssymb}
\usepackage[many]{tcolorbox}
\usepackage{enumitem}
\usepackage{lipsum}
\newcommand{\dint}{\displaystyle\int}
\newcommand{\dsum}{\displaystyle\sum}
\definecolor{myblue}{RGB}{0,163,243}

\newtcolorbox[auto counter,number within=section]{exo}[1][]{
  enhanced jigsaw,colback=white,colframe=myblue,coltitle=myblue,
  fonttitle=\bfseries\sffamily,
  sharp corners,
  detach title,
  leftrule=18mm,
  underlay unbroken and first={\node[below,text=white,font=\sffamily\bfseries,align=center]
    at ([xshift=-11mm,yshift=-1mm]interior.north west) {Exercice\\\thetcbcounter};},
  breakable,pad at break=1mm,
  #1,
  code={\ifdefempty{\tcbtitletext}{}{\tcbset{before upper={\tcbtitle\par\medskip}}}},
}


\begin{document}

\begin{exo}
\begin{enumerate}[wide=0pt, leftmargin=*]
\item Soit $\alpha >0$ , Montrer que $\smash[b]{\dint_{0}^{1}\dfrac{dt}{1+t^{\alpha }}=%
\dsum_{n=\,0}^{+\infty}\dfrac{(-1)^{n}}{1+n\alpha }}$ .
En déduire les sommes
\begin{equation*}
\sum_{n\,=1}^{+\infty }\dfrac{(-1)^{n}}{n}~;\overset{%
+\infty }{\underset{n\,=\,0}{\sum }}\dfrac{(-1)^{n}}{1+2n}\, \
\end{equation*}%
Indication : On pourra utiliser que \useshortskip
\[ \frac{1}{1+u}=\sum_{k\,=\,0}^{n}( -1)^{k}u^{k} + \frac{(-1)^{n+1}u^{n+1}}{1+u} \]

\item Calculer $\dsum _{n=\,0}^{+\infty }\dfrac{(-1)^{n}}{1+n}$ et $\dsum_{n=\,0}^{+\infty }\dfrac{(-1)^{n}}{1+2n}$.

\item On pose, pour $\alpha >0$, $R_{n}( \alpha) = \smashoperator{\dsum_ {k=n+1}^{+\infty}}\:\dfrac{(-1)^{k}}{1+k\alpha}$, montrer que :%
\begin{align*}
R_{n}( \alpha) & =( -1) ^{n+1}\int_{0}^{1}\dfrac{%
t^{\left( n+1\right) \alpha }}{1+t^{\alpha }}\,dt\\
 & =\dfrac{( -1)^{n+1}}{2( n+1) \alpha +2}+\dfrac{\alpha( -1) ^{n+1}}{%
( n+1) \alpha +1}\int_{0}^{1}\dfrac{t^{( n+2) \alpha }%
}{( 1+t^{\alpha }) ^{2}}\,dt
\end{align*}%
Montrer que $R_{n}( \alpha) \underset{+\infty }{\sim }\dfrac{( -1) ^{n+1}}{2( n+1) \alpha +2}.$

\item Étudier la nature de la série $\sum R_{n}\left( \alpha \right) .$
\end{enumerate}
\end{exo}

\end{document} 

Answer 3

改用align*环境：

\documentclass{article}
\usepackage{amsmath, amssymb}
\usepackage[many]{tcolorbox}
\usepackage{lipsum}
\DeclareMathOperator{\dint}{\displaystyle\int}
\DeclareMathOperator{\dsum}{\displaystyle\sum}
\definecolor{myblue}{RGB}{0,163,243}

\newtcolorbox[auto counter,number within=section]{exo}[1][]{
  enhanced jigsaw,colback=white,colframe=myblue,coltitle=myblue,
  fonttitle=\bfseries\sffamily,
  sharp corners,
  detach title,
  leftrule=18mm,
  underlay unbroken and first={\node[below,text=white,font=\sffamily\bfseries,align=center]
    at ([xshift=-11mm,yshift=-1mm]interior.north west) {Exercice\\\thetcbcounter};},
  breakable,pad at break=1mm,
  #1,
  code={\ifdefempty{\tcbtitletext}{}{\tcbset{before upper={\tcbtitle\par\medskip}}}},
}


\begin{document}

\begin{exo}
\begin{enumerate}
\item Soit $\alpha >0$ , Montrer que $\dint_{0}^{1}\dfrac{dt}{1+t^{\alpha }}=%
\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{1+n\alpha }$ .
En d\'{e}duire les sommes 
\begin{equation*}
\overset{+\infty }{\underset{n\,=1}{\sum }}\dfrac{(-1)^{n}}{n}~;\overset{
+\infty }{\underset{n\,=\,0}{\sum }}\dfrac{(-1)^{n}}{1+2n}
\end{equation*}%

Indication : On pourra utiliser que\textit{\ }$\dfrac{1}{1+u}=\overset{n}{%
\underset{k\,=\,0}{\dsum }}\left( -1\right) ^{k}u^{k}+\dfrac{\left(
-1\right) ^{n+1}u^{n+1}}{1+u}$

\item Calculer $\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{%
1+n}$ et $\overset{+\infty }{\underset{n=\,0}{\dsum }}\dfrac{(-1)^{n}}{1+2n}$

\item On pose , pour $\alpha >0$ , $R_{n}\left( \alpha \right) =\overset{%
+\infty }{\underset{k=n+1}{\dsum }}\dfrac{(-1)^{k}}{1+k\alpha }$ , montrer
que :%
\begin{align*}
R_{n}\left( \alpha \right)  &=\left( -1\right) ^{n+1}\int_{0}^{1}\dfrac{t^{\left( n+1\right) \alpha }}{1+t^{\alpha }}dt\\
~                           &=\dfrac{\left( -1\right)^{n+1}}{2\left( n+1\right) \alpha +2}+\dfrac{\alpha \left( -1\right) ^{n+1}}{
                            \left( n+1\right) \alpha +1}\int_{0}^{1}\dfrac{t^{\left( n+2\right) \alpha }}{\left( 1+t^{\alpha }\right) ^{2}}dt
\end{align*}
Montrer que $R_{n}\left( \alpha \right) \underset{+\infty }{\sim }\dfrac{%
\left( -1\right) ^{n+1}}{2\left( n+1\right) \alpha +2}.$

\item Etudier la nature de la s\'{e}rie $\sum R_{n}\left( \alpha \right) .$
\end{enumerate}
\end{exo}
\end{document}

Answer 4

我认为您不应该想加宽文本块。相反，请了解包align*的环境（在下面的示例中使用）amsmath。此外，您不应该过分依赖\left和\right。对于手头的文档，没有任何需要\left和\right实例。

另外，请养成书写的习惯，\dsum_{n=0}^{\infty}而不是\overset{n}{\underset{k\,=\,0}{\dsum }}。

\documentclass{article}
\usepackage{amsmath, amssymb}
\allowdisplaybreaks
\usepackage[many]{tcolorbox}
\usepackage{lipsum}
\usepackage[T1]{fontenc}   % to allow direct writing of "é"
\usepackage[french]{babel} % obey various French typographic criteria

\definecolor{myblue}{RGB}{0,163,243}
\newtcolorbox[auto counter,number within=section]{exo}[1][]{
  enhanced jigsaw,colback=white,colframe=myblue,coltitle=myblue,
  fonttitle=\bfseries\sffamily,
  sharp corners,
  detach title,
  leftrule=18mm,
  underlay unbroken and first= {\node[below,text=white,font=\sffamily\bfseries,align=center]
    at ([xshift=-11mm,yshift=-1mm]interior.north west) {Exercice\\\thetcbcounter};},
  breakable,pad at break=1mm,
  #1,
  code={\ifdefempty{\tcbtitletext}{}{\tcbset{before upper={\tcbtitle\par\medskip}}}},
}

%% Two new commands:
\providecommand\dsum{\displaystyle\sum}
\providecommand\dint{\displaystyle\int}

\begin{document}

\begin{exo}
\begin{enumerate}
\item Soit $\alpha >0$.
Montrer que $\dint_{0}^{1}\dfrac{dt}{1+t^{\alpha }}=
\dsum_{k=0}^{\infty}\dfrac{(-1)^{n}}{1+n\alpha}$.

En déduire les sommes 
$\dsum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n}$, 
$\dsum_{n=0}^{\infty}\dfrac{(-1)^{n}}{1+2n}$.

Indication: On pourra utiliser que
\[
\frac{1}{1+u}=\sum_{n=0}^{\infty}(-1)^{k}u^{k}+
\frac{(-1) ^{n+1}u^{n+1}}{1+u}\,.
\]

\item Calculer 
$\dsum_{n=0}^{\infty}\dfrac{(-1)^{n}}{1+n}$ et 
$\dsum_{n=0}^{\infty}\dfrac{(-1)^{n}}{1+2n}$\,.

\item On pose, pour $\alpha >0$, 
$R_{n}(\alpha) =\dsum_{k=n+1}^{\infty}\dfrac{(-1)^{k}}{1+k\alpha }$\,.

Montrer que:
\begin{align*}
R_{n}(\alpha) 
&=(-1) ^{n+1}\int_{0}^{1}\dfrac{t^{( n+1)\alpha }}{1+t^{\alpha }}\,dt \\
&=\dfrac{(-1)^{n+1}}{2(n+1) \alpha +2}+
  \dfrac{\alpha (-1)^{n+1}}{(n+1) \alpha +1}
   \int_{0}^{1}\dfrac{t^{(n+2) \alpha }}{( 1+t^{\alpha})^{2}}\,dt\,.
\end{align*}
Montrer que $R_{n}(\alpha) \underset{+\infty }{\sim }\dfrac{%
(-1) ^{n+1}}{2(n+1) \alpha +2}.$

\item Etudier la nature de la série 
$\dsum R_{n}(\alpha)$.
\end{enumerate}
\end{exo}
\end{document}