real analysis - Convergent sequence and $lim_limits{n to infty}n(a_{n+1}-a_n)=0$

Let $(a_n)_n$ be a real, convergent, monotonic sequence. Prove that if the limit $$\lim_{n \to \infty}n(a_{n+1}-a_n)$$ exists, then it equals $0$.

I tried to apply the Stolz-Cesaro theorem reciprocal:
$$\lim_{n\to \infty}n(a_{n+1}-a_n)=\lim_{n \to \infty} \frac{a_n}{1+\frac{1}{2}+\dots+\frac{1}{n-1}}=0$$ but I can't apply it since for $b_n=1+\frac{1}{2}+\dots+\frac{1}{n-1}$ we have $\lim_\limits{n \to \infty} \frac{b_{n+1}}{b_n}=1$. I also attempted the $\epsilon$ proof but my calculations didn't lead to anything useful.

Answer

Hint. You are on the right track. Note that $a_n$ is convergent to a finite limit and the harmonic series at the denominator is divergent. Therefore
$$0=\lim_{n \to \infty} \frac{a_n}{1+\frac{1}{2}+\dots+\frac{1}{n-1}}\stackrel{\text{SC}}{=}\lim_{n\to \infty}\frac{a_{n+1}-a_n}{\frac{1}{n}}=\lim_{n\to \infty}n(a_{n+1}-a_n).$$