Since $f$ is non-increasing, we have
$$
f(1) \ge \int_1^2 f(x)\, dx.
$$
Similarly,
$$
f(2) \ge \int_2^3 f(x) \,dx.
$$
In general, we have
$$
f(n) \ge \int_n^{n+1} f(x) \,dx,
$$
so
$$
s_n\ge\int_1^{n+1} f(x) \,dx.
$$
Taking the limit as $n\to\infty$, we have
Since
$$
s_n\ge\int_1^{n+1} f(x) \,dx,
$$
if the integral diverges, then the partial sum $s_n$ also goes to infinity, making the series
$\sum f(n)$ diverge.