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
| $$ \sum_{n=1}^\infty f(n) \ge \int_1^\infty f(x) \,dx$$ |
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.