How to prove $\frac{1}{n+1}$ is a Cauchy sequence

Question

I'm a little stumped by how I should go about proving that $\frac{1}{n+1}$ is a Cauchy sequence. I know that $\frac{1}{n}$ is a Cauchy sequence and understand the proof for that, but don't quite get how I can use that to show this is a Cauchy sequence.

I know that the definition of a Cauchy sequence is that for any sequence $a$, $\epsilon > 0$, and any $n,m > N$ for some $N \in \mathbb{N}$, $\left|a_n - a_m\right| < \epsilon$.

In this case, I would end up with $\left|\frac{1}{n+1} - \frac{1}{m+1}\right| < \varepsilon$ which I can use to find that $N$ would be $\frac{2}{\epsilon} - 1$, but I don't think this could work because it would be possible for our $N$ term to be negative. What is it I'm missing for this proof to work?

Answer 1

Hint $$|a_n-a_m|= \frac{1}{n+1}-\frac{1}{m+1} \leq \frac{1}{n+1}+\frac{1}{m+1} \leq \frac{2}{N+1} < \epsilon $$

Answer 2

Every convergent series in $\mathbb R$ is a Cauchy-sequence.

Let $a_n$ be a convergent series with limit $a \in \mathbb R$. Then there exist an $N \in \mathbb R$ such that $|a_n-a|\lt \frac{\epsilon}{2}$ and $|a_m-a|\lt \frac{\epsilon}{2}$ for all $n,m\gt N$

Hence $|a_n -a_m|=|a_n-a+a-a_m|\le|a_n-a|+|a_m-a|<\epsilon$

For your example: $a_n=\frac{1}{n+1}$ clearly converges to 0 and therefore is Cauchy-sequence

Answer 3

I'm a little stumped by how I should go about proving that $\frac{1}{n+1}$ is a Cauchy sequence. I know that $\frac{1}{n}$ is a Cauchy sequence and understand the proof for that, but don't quite get how I can use that to show this is a Cauchy sequence.

Well, $\frac{1}{n+1}$ is a subsequence of $\frac{1}{n}$. And a subsequence of a Cauchy sequence is also Cauchy.