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


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 \(N \in \mathbf{N}\) such that for any \(\epsilon > 0\), we have

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

This inequality is implying that the terms in the sequence are going to fall within the neighborhood of \(1\) (\(V_\epsilon(1)\)) for some value \(n \geq N\). In other words, all the terms starting at \(n \geq N\) will be within a radius of \(\epsilon\) around \(1\).

To find \(N\), we just need to solve the inequality for \(n\),

$$ \begin{align*} \frac{n+1}{n} - 1 &< \epsilon \\ \frac{1}{n} &< \epsilon \\ n &> \frac{1}{\epsilon}. \\ \end{align*} $$

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

Formal Proof

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

$$ \begin{align*} n &> \frac{1}{\epsilon} \\ \frac{1}{n} &< \epsilon \\ \frac{n+1}{n} - 1 &< \epsilon. \end{align*} $$

\(\frac{n+1}{n} > 1\) so we can re-write the last inequality as

$$ \begin{align*} | \frac{n+1}{n} - 1 | &< \epsilon. \end{align*} $$

And this is what we need to prove that the limit converges to 1 \(\blacksquare\).
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