**Lemma** Suppose that $f: [1, \infty) \to [0, \infty)$ is a non-negative [[Monotone Real Function|decreasing function]]. Then: 1. if $\int_{1}^n f(x) \, dx$ is [[Bounded Sequence|bounded]] then $\sum_{n=1}^{\infty} f(n) < \infty$ (i.e. it is [[Convergent Real Series|convergent]]); 2. if $\int_{1}^{n} f(x) \, dx$ is unbounded then $\sum_{n=1}^{\infty} f(n) = \infty$ **Proof** 1. Since $f$ is decreasing we have $f(n) \leq f(x) \quad \forall x\in [n-1, n]$since n is constant, integrating both sides between $n-1$ and $n$ gives $f(n) = f(n)\,[n-(n-1)]= \int_{n-1}^{n} f(n) \, dx \leq \int_{n-1}^{n} f(x) \, dx $taking the sums for $k=2$ to $n$ of both sides gives $\sum_{k=2}^{n}f(k) \leq \sum_{k=2}^{n} \int_{k-1}^{k} f(x) \, dx = \int_{1}^{n} f(x) \, dx $ ![[Bounding the sum from above using an integral.png|300]] Now if RHS is bounded, then the sum converges by [[Series with Non-Negative Terms Converges Iff Partial Sums Are Bounded Above|lemma 3.4]]. 2. Conversely, note $f(n) \geq f(x) \quad \forall x \in [n,n+1]$integrating both sides between $n$ and $n+1$ we obtain $f(n) = \int_{n}^{n+1} f(x) \, dx \geq \int_{n}^{n+1} f(x) \, dx $Therefore $\sum_{k=1}^{n} f(k) \geq \int_{1}^{n+1} f(x) \, dx $If the RHS is unbounded, then the LHS diverges by [[Series with Non-Negative Terms Converges Iff Partial Sums Are Bounded Above|lemma 3.4]]. ### Applications - [[p-series converges when p > 1]]. - [[Harmonic Numbers]].