Prove that the sequence is Cauchy

Question

This is problem 17, page 55, from section 1.2: Cauchy Sequences in the textbook Introduction to Analysis, Fifth Edition, by Edward D. Gaughan.

17. Prove that the sequence $\left\{\frac{2n-1}{n}\right\}_{n=1}^{\infty}$ is Cauchy.

Answer 1

Proof: Because every convergent sequence is Cauchy, it suffices to prove that the sequence converges.

Choose $\epsilon>0$. Let $N$ be an integer greater than $\frac{1}{\epsilon}$. Then, for $n\geq N$, we have $\frac{1}{n} \leq \frac{1}{N} < \epsilon$.

Making the educated guess that the sequence converges to 2, for $n\geq N$, we have

$|\frac{2n-1}{n} - 2| = |2 - \frac{1}{n} - 2 | = |\frac{1}{n}|\leq\frac{1}{N}<\epsilon$

Thus, the sequence converges to 2, and since every convergent sequence is Cauchy, this concludes the proof.

I welcome any corrections or critique.