solving integral inequality

Question

How can I show that

$0 \le \int_n^{n+1}\frac{1}{n}-\frac{1}{x}dx\le \frac{1}{n}-\frac{1}{n+1}$

I think I need to take the log to solve this, but I'm not quite sure.

Answer 1

Hint:

Just note that

• $\int_n^{n+1}c\;\;dx = c$
• $n \leq x \leq n+1 \Leftrightarrow \frac{1}{n} \geq \frac{1}{x} \geq \frac{1}{n+1}$ for $n \in \mathbb{N}$

Answer 2

It should be clear that $0 \le \int_n^{n+1}(\frac{1}{n}-\frac{1}{x})dx$.

The mean value theorem for integrals shows that there is $t \in [n,n+1]$ such that

$ \int_n^{n+1}(\frac{1}{n}-\frac{1}{x})dx= \frac{1}{n}-\frac{1}{t}$. Since $t \le n+1$, the result follows.