calculus - Prove $intlimits_1^infty f(x) , dx

Theorem: Let $f$ be a positive function, continuous and decreasing in an interval $[1,\infty]$. If

$$\int_1^\infty f(x)\,dx<\infty$$

therefore we have:

$$\sum_{i=1}^\infty f(i)<\infty$$

and reciprocally.

By the definition of $\displaystyle\int_1^\infty f(x)\,dx=\lim_{\Delta x\to 0} \sum_{i=1}^\infty f(x)\,\Delta x$

As the $\lim_{\Delta x}\Delta x=dx$ is constantly infinitesimally small, we conclude that if $\int\limits_1^\infty f(x) \, dx<\infty$, then $\sum\limits_{i=1}^\infty f(i)<\infty$ is true.

Questions:

Do you think this prove is right? Can I treat $\lim_{\Delta x}\Delta x$ as constant?

Answer

\begin{align}
& \int_1^\infty f(x)\,dx = \sum_{i=1}^\infty \int_i^{i+1} f(x)\,dx \le \sum_{i=1}^\infty \int_i^{i+1} f(i)\, dx = \sum_{i=1}^\infty f(i) <\infty. \\
& \text{This instance of “$\le$'' is true because $f$ is decreasing.} \\[20pt]
& \sum_{i=1}^\infty f(i) \le \sum_{i=2}^\infty f(i) = \sum_{i=2}^\infty \int_{i-1}^i f(i)\,dx \le \sum_{i=1}^\infty \int_{i-1}^i f(x)\,dx = \int_1^\infty f(x)\,dx < \infty. \\
& \text{The second “$\le$'' on this line is true because $f$ is decreasing.}
\end{align}