$\displaystyle
{a}_{n}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}
=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{n} }\frac{1}{k}
<\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{{k}^{3/2}}
$
Let $\displaystyle x,t\in \left(a,b \right)$.
By the mean value theorem of derivatives:
$\displaystyle f(x)-f(t)=(x-t)f'(\lambda )$
where $\displaystyle \lambda \in (x,t)$. Hence:
$\displaystyle \left| f(x)-f(t)\right|\leq M\left| x-t\right|$
Now for any $\displaystyle \epsilon >0$, we can put
$\displaystyle \delta =\frac{\epsilon }{M}$
so that whenever $\displaystyle \left| x-t\right|<\delta $, we will have $\displaystyle \left| f(x)-f(t)\right|<\epsilon $. This proves that $\displaystyle f(x)$ is uniformly continuous.