Answer
See proof below.
Work Step by Step
Let $0\lt M \lt 1.$
We want to find N such that
$n \gt N\displaystyle \Rightarrow\frac{n}{n+1} \gt M$
\begin{align*}
\displaystyle \frac{N}{N+1}&=M \\
N&=M(N+1) \\
N&=MN+M \\
N-MN&=M \\
N(1-M)& =M \\
N& =\displaystyle \frac{M}{1-M} \end{align*}
For $n \gt N =\displaystyle \frac{M}{1-M} ,$
\begin{align*}
n& \displaystyle \gt \frac{M}{1-M}\tag{multiply}\\
n-nM& \gt M\\
n& \gt M+nM\\
n& \gt M(1+n) \\
\displaystyle \frac{n}{n+1}&\gt M \end{align*}
which is what was needed to be shown.
