Q:

My proof of ratio lemma for sequences tending to inf.

My proof of $(b_n)$ is a sequence of positive numbers. Support there exists $z\gt 1$ s.t. $b_{n+1}/b_n\ge z\forall n$ .Show $(b_n)\rightarrow \infty$ 

If $l\gt 1$ s.t. $b_{n+1}/b_n\ge z\forall n$ then $b_{n+1}/b_n\ge z\gt1$. So $b_{n+1}\ge zb_n\gt b_n\implies b_{n+1}\gt b_n\forall n$. Since this is a definition of a strictly increasing sequence we get that $(b_n)\rightarrow \infty$.

Any comments are appreciated.

A:

Your proof is wrong. There are strictly increasing sequences that do not converge to $+\infty$. An example of such a sequence is $-1/n \nearrow 0$.
One way to do this properly is to note that for all $n \geq 1$
$$b_{n} \geq zb_{n-1} \geq z^2 b_{n-2} \geq \dots \geq z^{n-1} b_1$$
Since $\lim_{n \to \infty} z^{n-1} b_1 = + \infty$ (since $z > 1$), it follows that $\lim_{n \to \infty} b_n = + \infty$ by the squeeze theorem.

