I have to prove the following inequality:
$1/(n+1) < \int_n^{n+1} 1/t$ $dt$ $<1/n$
I thought it would be easier to attack this via integration, so I get:
$1/(n+1) <$ log $(n+1)-$ log$(n)<1/n$
At this point I tried to use induction, but the solution is still not clear to me.
Thanks a lot!
Hint: Draw a picture. On the interval $[n,n+1]$, our function $\frac{1}{t}$ is $\le \frac{1}{n}$, and $\ge \frac{1}{n+1}$.
So the area under the curve $\frac{1}{t}$, and above the $t$-axis, from $t=n$ to $t=n+1$, is less than the area of a rectangle with base $1$ and height $\frac{1}{n}$, and greater than the area of a rectangle with base $1$ and height $\frac{1}{n+1}$.
From the above geometric argument, it follows that $\frac{1}{n+1}\lt \log(n+1)-\log n\lt \frac{1}{n}$.