From the definition of convergence , for every $\u03f5>0$ there is $N(\u03f5)\in \mathbb{N}$ such that $(\forall )n\ge N(\u03f5)$ , we have :

Because ${b}_{n}$ is strictly increasing^{} we can multiply the last equation with ${b}_{n+1}-{b}_{n}$ to get :

Let $k>N(\u03f5)$ be a natural number^{} . Summing the last relation^{} we get :

Divide the last relation by ${b}_{k+1}>0$ to get :

This means that there is some $K$ such that for $k\ge K$ we have :

(since the other terms who were left out converge^{} to 0)