0

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 $\left(\frac{1}{\lfloor x\rfloor}\right)$?

Zain Patel
  • 16,802
Mekanik
  • 1,761
  • You have $\lfloor x \rfloor \leqslant x < \lfloor x\rfloor + 1$, so for $a \geqslant 0$, you have $$\frac{1}{\lfloor x+1\rfloor^a} < \frac{1}{x^a} \leqslant \frac{1}{\lfloor x\rfloor^a}.$$ – Daniel Fischer Jul 05 '15 at 11:15

1 Answers1

1

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.

Reactant
  • 677