Proof of the comparison of a series and an integral

Proof

We compare $f$ with a constant on each interval of length $1$.

1. If $n = 0$, all three expressions are $0$: the sums are empty and the interval $[a, a]$ is a point. From now on let $n \ge 1$.
2. Since $f$ is decreasing, it is monotone. So $f$ is Riemann integrable on $[a, a+n]$ and on each subinterval, in the sense of the Riemann integral.
3. By additivity of the integral over intervals, applied $n - 1$ times, $$\int_a^{a+n} f(x)\, dx = \sum_{i=1}^{n} \int_{a+i-1}^{a+i} f(x)\, dx.$$
4. Let $1 \le i \le n$ and $x \in [a+i-1, a+i]$. Since $f$ is decreasing, $f(a+i) \le f(x) \le f(a+i-1)$.
5. The integral of a constant $c$ over an interval of length $1$ is $c$. By monotonicity of the integral and step 4, $$f(a+i) \le \int_{a+i-1}^{a+i} f(x)\, dx \le f(a+i-1).$$
6. Add the inequalities of step 5 over $i = 1, \dots, n$ and use step 3. This gives $$\sum_{i=1}^{n} f(a+i) \le \int_a^{a+n} f(x)\, dx \le \sum_{i=1}^{n} f(a+i-1) = \sum_{i=0}^{n-1} f(a+i).$$