I need to prove that {(2n+1/n} is Cauchy. I know that the theorem states that if |s_m-s_n| < epsilon for n>=N, then the sequence is Cauchy, but I am not sure how to start.
Note that $\displaystyle \frac{2n+1}{n}=2+\frac{1}{n}$. I claim this sequence converges to $\displaystyle 2$.
$\displaystyle \left|2+\frac{1}{n}-2\right|=\left|\frac{1}{n}\right|$
If we let $\displaystyle N=\frac{1}{\epsilon}$, $\displaystyle n>N$ implies that $\displaystyle \frac{1}{n}<\epsilon$, so this sequence converges. If a sequence converges, then it is Cauchy* (but not necessarily the other way around).
*If this is not obvious, I can provide a proof.
Yes, it is absolutely obvious, and thank you. We don't get much in class except the proofs of the theorems, no examples with actual numbers in them. I am sometimes slow to relate one to the other! I will use these calculations in formulating a proof and, if need be, as for further help, if that is ok. Thank you very much!
Showing convergence rigorously makes use of $\displaystyle \epsilon$-proofs. A sequence $\displaystyle a_n$ converges to a point $\displaystyle a$ iff for all $\displaystyle \epsilon>0$, $\displaystyle \exists~N$ such that $\displaystyle n>N$ implies $\displaystyle |a_n-a|<\epsilon$, and a sequence $\displaystyle a_n$ is Cauchy iff for all $\displaystyle \epsilon>0$, $\displaystyle \exists~N$ such that $\displaystyle m,n>N$ implies $\displaystyle |a_m-a_n|<\epsilon$.