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

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