How do you prove that $\frac{1}{n+1}\le\int_{n}^{n+1}\frac{1}{x}dx\le\frac{1}{n}$?

Question

I tried to simplify things to see what I could do:

$\frac{1}{n+1}\le\int_{n}^{n+1}\frac{1}{x}dx\le\frac{1}{n}$ $\implies\frac{1}{n+1}\le ln(n+1)-ln(n)\le\frac{1}{n}$

However, after this, I'm in a bit of a rut.

I then tried to say that

$\frac{1}{n+1}\le ln(n+1)-ln(n)$

and

$ln(n+1)-ln(n)\le\frac{1}{n}$

But that didn't really help either. I'm assuming I'm going the wrong direction; how should I go about this problem? Thanks!

edit: I should add that this is to be proved only for $n \ge 1$.

You were overthinking it. The function $1/x$ is decreasing on interval $[n,n+1]$ of length one. — hardmath, Jan 07 '19 at 04:27

Shubham Johri · Accepted Answer · 2019-01-07T04:14:32.723

6

For $x\in[n,n+1],n\in\Bbb N$, we have $$\frac1{n+1}\le\frac1x\le\frac1n\\\int_n^{n+1}\frac1{n+1}dx\le\int_n^{n+1}\frac1xdx\le\int_n^{n+1}\frac1ndx$$

edited Jan 07 '19 at 04:14

answered Jan 07 '19 at 04:09

Shubham Johri

I'm sorry, could you elaborate? – iaskdumbstuff Jan 07 '19 at 04:12
@user494405 See the edit. – Shubham Johri Jan 07 '19 at 04:15
Oh wow, that was so simple. Thanks for the help! – iaskdumbstuff Jan 07 '19 at 04:21

score 1 · Answer 2 · answered Jan 07 '19 at 06:10

1

By the mean value theorem for integrals, there is $a \in [n,n+1]$ such that

$\int_n^{n+1}\frac1xdx=\frac{1}{a}(n+1-n)=\frac{1}{a}$.

answered Jan 07 '19 at 06:10

Fred

2 Answers2