The first equation below features the frational term shifted down so that the fraction bar is at the same height as the lower horizontal stroke of the summation symbol. The second equation features a more standard look.
Speaking for myself, I'd go with the look of the second equation.
\documentclass{article}
\usepackage{amsmath,amsfonts}
\DeclareMathOperator{\1}{\mathbf{1}} % educated guess...
\newcommand\e{\mathrm{e}}
\begin{document}
fractional term shifted down:
\[
\e^{-\xi_1^{} Y_1^N}
= \exp\biggl(- \lower1.6ex\hbox{$\dfrac{\xi_1}{N}$}
\sum_{n\geq1}{}V_n^1
\e^{-s_n} \1_{\{ s_n \leq T_1\}}\biggr)
\]
\bigskip
normal look:
\[
\e^{-\xi_1^{} Y_1^N}
= \exp\biggl(- \frac{\xi_1}{N}
\sum_{n\geq1}{}V_n^1
\e^{-s_n} \1_{\{ s_n \leq T_1\}}\biggr)
\]
\end{document}
Addendum: Judging by the comment the OP left, it may be that the following look -- placing the limit of the summation to the side rather than below the summation symbol -- is really what is supposed to be achieved.
\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator{\1}{\mathbf{1}}
\newcommand\e{\mathrm{e}}
\begin{document}
\[
\e^{-\xi_1^{} Y_1^N}
= \exp\Bigl(- \frac{\xi_1}{N}
\sum\nolimits_{n\geq1}V_n^1
\e^{-s_n} \1_{\{ s_n \leq T_1\}}\Bigr)
\]
\end{document}
\1
macro defined?\1
is defined as\def\1{\mathbbm{1}}
Looking at your answer I realize that in fact what I "needed" was to shift up the Summation, rather than shifting sown the fraction, and it is only to fit what I do when handwriting in paper, thought it would be more tidy... If it gets similar to the first way; I'd rather keep it like it is now, too. Thank you for your help!\nolimits
immediately after\sum
. I've provided an addendum to my answer to show the resulting look.