Prove that \(\lim\big(\frac{n+1}{n}\big)= 1\).

For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template, see this.

Problem Discussion

We want to prove that \(\lim\big(\frac{n+1}{n}\big) = 1\). To do so we need to find \(N \in \mathbb{N}\) such that for any \(\epsilon > 0\), the tail of the sequence or the terms of the sequence end up in

$$ \begin{align*} \frac{n+1}{n} - 1 \in (1-\epsilon, 1+\epsilon) \quad \text{whenever $n \geq N$.} \end{align*} $$

In other words

$$ \begin{align*} \big\lvert \frac{n+1}{n} - 1 \big\rvert < \epsilon \quad \text{whenever $n \geq N$.} \end{align*} $$

We solve this by working backgrounds to derive what \(N\) needs to be in order for the sequence terms to fall within the epsilon neighborhood of \(1\). In other words, all the terms starting at \(n \geq N\) will be within a radius of \(\epsilon\) around \(1\). Solving for \(n\)

$$ \begin{align*} \big\lvert \frac{n+1}{n} - 1 \big\rvert &< \epsilon \\ \big\lvert 1 + \frac{1}{n} - 1 \big\rvert &< \epsilon \\ \big\lvert \frac{1}{n} \big\rvert &< \epsilon \\ \frac{1}{n} &< \epsilon \quad \text{($n$ is positive)} \\ n &> \frac{1}{\epsilon}. \\ \end{align*} $$

From this we see that we want \(n\) to be greater than \(1/\epsilon\) in order to make the inequality works.

Formal Proof

Let \(\epsilon > 0\) be arbitrary. Choose an integer \(N\) satisfying \(N > \frac{1}{\epsilon}\). We now verify that this choice is appropriate. Let \(n \geq N\). Then,

$$ \begin{align*} \big\lvert \frac{n+1}{n} - 1 \big\rvert &= \big\lvert 1 + \frac{1}{n} - 1 \big\rvert \\ &= \big\lvert \frac{1}{n} \big\rvert \\ &= \frac{1}{n} \quad \text{($n$ is positive)} \\ &\leq \frac{1}{N} \quad \text{(since $n \geq N$)} \\ &< \epsilon. \end{align*} $$

as desired. \(\blacksquare\)

References