10
\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}

enter image description here

  • Welcome to TeX.SX! Could you please make your code compilable? How are \dint, \dsum defined? – leandriis Jun 30 at 20:15
  • 3
    Welcome to tex.sx. A much better approach would be to break the overlong display. You are using amsmath, so the multline environment would work. Just before the second equals sign, insert \\ Details are in the amsmath user guide, texdoc amsldoc. – barbara beeton Jun 30 at 20:16
  • Thank you , I'm bigginer in latex, can you please write what can i change. – mustapha saadaoui Jun 30 at 20:20
7

For example you can use environment split to divide the long equation at the = signs like (see the added code marked with <======):

\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*}%

In your code are several errors I ignore because we do not know how you defined the related commands like \dint. If you add the definitions to your question I can update my answer.

Please see the following 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}

and its result:

enter image description here

5

I would suggest splitting one equation instead, and removing the many unnecessary \left \right pairs. Unrelated: I don't see why you didn't type your accented letters directly on the keyboard, all the more so as all modern TeX editors understand utf8. I added some improvements with enumitem, mathtools and 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} 

enter image description here

4

Use the align* environment instead:

Output

\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}
4

You might be interested in the package layout.

The pack­age de­fines a com­mand \lay­out, which will draw nice pictures showing a sum­mary of the lay­out of the cur­rent doc­u­ment.

Beware in mind:
You don't need this package for setting/adjusting margins.
But it is useful for getting a visual impression of how the layout of the current document is formed due to setting values for the various layout parameters of the standard document classes.

A few remarks about horizontal adjustments:

Usually \hoffset and \voffset are 0 and these values should not be changed as these are intended for horizontally and vertically shifting pages in case this is needed for compensating undesired horizontal and vertical shiftings done by some printing devices.

The left margin both for all pages of one-side-documents and for right-hand-pages of two-side-documents is:
1in+\hoffset+\oddsidemargin.

The left margin for left-hand-pages of two-side-documents is:
1in+\hoffset+\evensidemargin.

You cannot/do not need to specify right margins as they are specified indirectly by specifying left margins and the width of the text and the width of the paper.

With two-side-documents the outer margin of a page shall usually be twice as wide as the inner margin of that page.

Thus with two-side-documents the left margin of right hand pages shall usually be half as wide as the left margin of left hand pages.

If you wish to change/adjust margins for single pages only, you can do this by making sure that the material which forms the previous page is shipped out and then placing the material for the pages with changed/adjusted margins into a local scope on its own wherein as first thing the page layout parameters are set in terms of non-global assignments and wherein as last thing it is ensured that the material that shall form these pages is shipped out. I mentioned that you need to do that in terms of non-global assignments because LaTeX's \setlength and \addtolength will perform global assignments. In such situationy you can, e.g., use plain-TeX's \advance .. by .. instead.

You can do something like this:

\documentclass[a4paper]{article}
\usepackage{amsmath, amssymb}
\usepackage[many]{tcolorbox}
%\usepackage{lipsum}
\usepackage{layout}

\setlength{\oddsidemargin}{2.5cm}%
\setlength{\evensidemargin}{\oddsidemargin}%
\begingroup\makeatletter\@firstofone{%
  \endgroup
  \if@twoside
    \setlength{\oddsidemargin}{.5\oddsidemargin}%
  \fi
}%
\setlength{\textwidth}{\paperwidth}%
\addtolength{\textwidth}{-\oddsidemargin}%
\addtolength{\textwidth}{-\evensidemargin}%
\setlength{\marginparwidth}{.625\evensidemargin}%
\setlength{\marginparsep}{.15\evensidemargin}%
\addtolength{\oddsidemargin}{-1in}%
\addtolength{\evensidemargin}{-1in}%

\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}}}},
}

%\def\dint{\mathop{\displaystyle \int}}%
%\def\dsum{\mathop{\displaystyle \sum }}%

\begin{document}

\layout\newpage

\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}

enter image description here

enter image description here

3

I believe you shouldn't want to widen the textblock. Instad, do learn about the align* environment -- used in the example below -- of the amsmath package. In addition, you shouldn't rely on \left and \right as much. For the document at hand, none of the \left and \right instances are needed.

Also, do make a habit of writing \dsum_{n=0}^{\infty} instead of \overset{n}{\underset{k\,=\,0}{\dsum }}.

enter image description here

\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}
  • you can change \dfrac by \displaystyle\frac and \dsum by \displaystyle\sum – mustapha saadaoui Jun 30 at 20:55
  • @mustaphasaadaoui - The amsmath package provides a definition for \dfrac; hence, no need to define this instruction again. – Mico Jun 30 at 21:15
1

If you want to globally decrease the marigins you might want to use the geometry package. If you then slightly change the width of the left blue bar and adjsut the indentation of the items in the enumerate list using the enumitem package, the equation can fit into the textwidth without splitting it into two lines. (Since there was no definition for \dsum and \dint given in the question, I just used \sum and \intrespectively):

enter image description here

\documentclass[draft]{article}
\usepackage{geometry}
\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=16mm, %<-----------
  underlay unbroken and first={\node[below,text=white,font=\sffamily\bfseries,align=center]
    at ([xshift=-8mm,yshift=-1mm]interior.north west) {Exercice\\\thetcbcounter};}, % <-----------
  breakable,pad at break=1mm,
  #1,
  code={\ifdefempty{\tcbtitletext}{}{\tcbset{before upper={\tcbtitle\par\medskip}}}},
}


\newcommand{\dint}{\int}
\newcommand{\dsum}{\sum}

\usepackage{enumitem}
\setlist[enumerate]{leftmargin=4pt} %<-----------

\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}

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