Let $0 < b_n ≤ b_{n+1}$ for all $n$. Prove that $\lim b_n^{-1} = 0$ if and only if $b_n$ is an unbounded sequence
I know that if $b_n$ is bounded either it has a real numbered limit or it does not have a limit. So if it has $\lim b_n^{-1} = 0$ cannot be true. But I don't know how to prove this or what happens when it does not have a limit. Can you help me please ?
Suppose $\displaystyle \lim_{n \to \infty} \frac{1}{b_n} = 0$. Then for all $\varepsilon > 0$, there exists $N$ such that if $n > N$, then $\displaystyle \frac{1}{b_n} < \varepsilon$ which means $\displaystyle b_n > \frac{1}{\varepsilon}$. We can choose $\varepsilon$ to be as small as we want, making $b_n$ as big as we want, hence it's unbounded.
Suppose $b_n$ is an unbounded sequence, then for all $M > 0$, there exists $N$ such that $b_N > M$, but since $b_{N + 1} \geq b_N > M$ by hypothesis, if $n > N$, then $\displaystyle \frac{1}{b_n} < \frac{1}{M} = \varepsilon$. Conclude by definition that $\displaystyle \lim_{n \to \infty} \frac{1}{b_n} = 0$