Prove $\displaystyle \dfrac{n}{n + 1} > M$ whenever $n > N$ for some integer $N$

Question

Q: If $M$ is a positive number less than 1. Prove the terms in $\displaystyle \left\{\dfrac{n}{n + 1}\right\}_{n = 1}^{\infty}$ exceed $M$ for sufficiently large $n$; that is, prove $\displaystyle \dfrac{n}{n + 1} > M$ whenever $n > N$ for some integer $N$.

Not exactly sure how to prove this but here is my attempt so far.

By the definition of the limit, $\displaystyle \lim_{n \to \infty} \dfrac{n}{n + 1} > M$, we have $\forall M\in(0, 1)\exists N\in\mathbb{N}$ such that if \begin{equation*} n > N \hspace{1cm} \rightarrow \hspace{1cm} a_n > M \end{equation*}

If $a_n = \dfrac{n}{n + 1}$, then we want to show that it exceeds $M$ as $n$ gets larger.

This is the part where I am kind of lost. I am not exactly sure how to proceed from here. Some tips or advice would be useful. Thanks

Answer 1

Hint: $\frac{n}{n+1} = 1-\frac{1}{n+1}$. Fix an arbitrary $M < 1$ and write $M = 1 - \epsilon$ for some $\epsilon > 0$. Using archimedian property...

Answer 2

Both $M$ and $1-M$ are positive

if $$n > \frac{M}{1-M},$$ we see $$ n+1 > \frac{1}{1-M} $$ $$ \frac{1}{n+1} < 1-M $$ $$ -\frac{1}{n+1} > M-1 $$ $$ 1-\frac{1}{n+1} > M $$ $$ \frac{n}{n+1} > M $$

Answer 3

$\frac n{n+1} > M\iff n > M(n+1)\iff $

$n> Mn + M \iff n-Mn = n(1-M) > M$.

As $M < 1$ then $1- M > 0$ so

$n(1-M) > M \iff n \ge \frac M{1-M}$

So if $N$ is any integer $\ge \frac M{1-M}$ you get your result.

$n > N \ge \frac 1{1-M} \implies \frac n{n+1} > M$.