41301J6O.tex

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{enumerate}
\item \dots
\item \dots
\item \dots
\item \dots
\item \dots
\item \dots
\item \dots
\item Now calculate the logarithm of the $n$th term in the sequence, i.e. $\log a_n$.
  
  Use the series representation of the logarithm to simplify this answer:
  \[ \log (1 + x) = -\sum_{k=1}^\infty \frac{(-1)^k}{k} x^k. \]
  What do you conclude?
  \newcommand\frexp[1]{\frac{(-1)^{#1}}{#1} \cdot \frac{#1}{n^{#1}}}
  \begin{align*}
     \log (1 + \frac{1}{n})^n
     &= - n \cdot \log (1 + \frac{1}{n}) \\
     &= - n \cdot (-\sum_{k=1}^\infty \frac{(-1)^k}{k} (\frac{1}{n})^k) \\
     &= - n \cdot (\sum_{k=1}^\infty \frac{(-1)^k}{k} (\frac{1}{n})^k) \\
     &= - n \left[ \underbrace{\frexp{1}}_{n=1} + \underbrace{\frexp{2}}_{n=2} + \cdots \right] \\
     &= \dots
  \end{align*}
\end{enumerate}
\end{document}