Let $p=1+\frac{1}{\sqrt 2}+\cdots\frac{1}{\sqrt {120}}$ and $q=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots\frac{1}{\sqrt {121}}$ then

Here is one way:

Consider trapeziums formed by the points: $$(a,0),(a,f(a)),(a+1,0),(a+1,f(a+1))$$ for $a=1,2,....120$, and note that the sum of areas of these trapeziums is greater than $\int_1^{121} f(x) dx$. Also note that the sum of areas of the trapeziums is nothing but $\frac{p+q}{2}$. And the conclusion follows.

Hint:
Utilize approximation $$S(n)=\sum_{k=1}^n\frac{1}{\sqrt k} \sim \frac{4n+1}{2\sqrt n}-\sqrt 2 $$