Since the limit of the sequence $b^n$ where $0The question is in the title. For a fixed number $b \in \mathbb{R}$ where $0<b<1$,  I can prove that the sequence $b^n$ converges to zero. 
Proof 
Choose $K(\epsilon) = \lfloor \frac{\ln \epsilon}{\ln b} \rfloor + 1$. Then for all $n \geq K$ we have $$ \ln b^n < \ln \epsilon \implies \left|b^n - 0 \right| < \epsilon $$
So, can I say something like... clearly we have ,  $ 0 < n/(n+1) < 1, \forall n\in \mathbb{N}$ so set $b = n/(n+1)$ then 
$$ \lim \left(\frac{n}{n+1} \right)^n = 0$$

write your term in the form
$$\frac{1}{\left(\frac{n+1}{n}\right)^n}=\frac{1}{\left(1+\frac{1}{n}\right)^n}$$
