Let $f$ be Riemann-integrable in the interval $[1,\infty)$ and let for all $x \geq 1$ $f(x) \geq \left(\frac{1}{\lfloor x\rfloor}\right)^\alpha$. Then $\alpha > 1$. How to prove this statement?
Ok for example for $\alpha = 1$ I know that $1/x$ is not integrable in the above interval. How do I compare this with the integral of $\left(\frac{1}{\lfloor x\rfloor}\right)$?
When $x \geq 1$ $$f(x) \geq \left(\frac{1}{\lfloor x \rfloor} \right)^\alpha > 0$$
We have
$$\int_{1}^{\infty} f(x) dx \geq \int_{1}^{\infty} \left(\frac{1}{\lfloor x \rfloor} \right)^\alpha dx = \sum_{k = 1}^{\infty} \frac{1}{k^\alpha} $$
If $\alpha \leq 1$
$$\sum_{k = 1}^{\infty} \frac{1}{k^\alpha}$$ diverges so $$\int_{1}^{\infty} f(x) dx $$ diverges also.