1. ## Cauchy Sequence Proof

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.

2. Originally Posted by tarheelborn
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.
If the sequence is $s_n=\left(2n+\frac{1}{n}\right)$ then it is not a Cauchy sequence.

3. I am so sorry; I missed a closing parentheses. It should be {(2n+1)/n}.

4. Originally Posted by tarheelborn
I am so sorry; I missed a closing parentheses. It should be {(2n+1)/n}.
Well $s_n=\frac{2n+1}{n}=2+\frac{1}{n}$.
So $s_n-s_m=\left(2+\frac{1}{n}\right)-\left(2+\frac{1}{m}\right)=\frac{1}{n}-\frac{1}{m}$.

Is $\frac{1}{n}$ a Cauchy sequence?

5. Originally Posted by tarheelborn
I am so sorry; I missed a closing parentheses. It should be {(2n+1)/n}.
Note that $\frac{2n+1}{n}=2+\frac{1}{n}$. I claim this sequence converges to $2$.

$\left|2+\frac{1}{n}-2\right|=\left|\frac{1}{n}\right|$

If we let $N=\frac{1}{\epsilon}$, $n>N$ implies that $\frac{1}{n}<\epsilon$, 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.

6. 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!

7. I got stuck quickly. Do I need to do the epsilon proof or will I be able to simply show convergence? My professor is all about the epsilon proofs, but she didn't say it had to be that way for this problem.

8. Showing convergence rigorously makes use of $\epsilon$-proofs. A sequence $a_n$ converges to a point $a$ iff for all $\epsilon>0$, $\exists~N$ such that $n>N$ implies $|a_n-a|<\epsilon$, and a sequence $a_n$ is Cauchy iff for all $\epsilon>0$, $\exists~N$ such that $m,n>N$ implies $|a_m-a_n|<\epsilon$.