Does the difference of a monotonic function and the floor of its argument always converge?

Question

There are several similar questions already posted, such as (Integral of the difference between a function and its floor), but none seem to address my question in the general case.

Consider a monotonically decreasing f(x) which has no singularities over the integrated region. My question is about expressions of the form:

$$\int_n^\infty f(x) - f(\lfloor x \rfloor) \ dx $$

For example, if f(x) = $\frac{1}{x}$ and $n=1$ then the above expression converges to $-\gamma$ : the Euler–Mascheroni constant. Will this expression converge generally for these type of functions or are there counterexamples?

Answer 1

If $f$ is not bounded below, it does not need to converge. For example, try $f(t) = -t$.

If it is bounded below (and thus has a limit as $n \to \infty$), note that $$0 \ge \int_n^{n+1} (f(x) - f(n))\; dx \ge \int_n^{n+1} (f(n+1) - f(n))\; dx = f(n+1) - f(n)$$ so that $$ 0 \ge \int_n^\infty (f(x) - f(n))\; dx \ge \sum_{k=n}^\infty (f(k+1) - f(k)) = \lim_{k \to \infty} f(k) - f(n)$$