(related to Proposition: Integral Test for Convergence)

By hypothesis, $N\in\mathbb N$ is a non-negativ natural number ($N\ge 1$) and $f:[N,\infty)\to\mathbb R$ is a monotonically decreasing and non-negative function.

- Assume, the infinite series $$\sum_{n=N}^\infty f(n)$$ is convergent.
- By definition of monotonically decreasing real-valued function for all natural numbers $n\ge N$ we have that if $n-1\ge x\ge n$ then $f(n)\le x\le f(n-1).$
- With $s:=\sup\{f(x)\mid n-1\ge x\ge n\}$ we get the estimation $$f(n)\le\int_{n-1}^n f(x)dx\le s\cdot 1\le f(n-1).$$
- The sum for $n=N+1,\ldots M$ yields $$\sum_{n=N+1}^Mf(n)\le\int_{N}^M f(x)dx\le \sum_{n=N}^{M-1}f(n).\label{eq:E20282}\tag{1}$$
- By assumption, the right side of the inequation $(\ref{eq:E20282})$ converges to a constant for $M\to\infty$. Therefore, for all $M\ge N$, the real sequence $(r_M)_{M\in\mathbb N}$ of Riemann integrals $$r_M:=\int_{N}^M f(x)dx$$ is bounded above by this constant.
- Moreover, the left side of the inequation $(\ref{eq:E20282})$ shows that the sequence $(r_M)_{M\in\mathbb N}$ is monotonically increasing, since all $f(n)$ are by definition non-negative.
- Since every bounded monotonic sequence is convergent it follows that the improper integral $$\int_{N}^\infty f(x)dx$$ is convergent.

- Assume, $\int_{N}^\infty f(x)dx$ is convergent.
- Then, the left side of the inequation $(\ref{eq:E20282})$ shows that the sequence $(s_M)_{M\in\mathbb N}$ of the partial sums $$s_M:=\sum_{n=N+1}^Mf(n)$$ is bounded above.
- Moreover, it is monotonically increasing, since all $f(n)$ are by definition non-negative.
- It follows that it is a convergent real sequence.
- By definition of convergent real series, the series $\sum_{N+1}^\infty f(n)$ is convergent and so is the series $$\sum_{N}^\infty f(n).$$∎

**Forster Otto**: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983