Show $\sum_{k=1}^{n}\frac{1}{k}\sim \ln(n)$ there is an example of how we apply Integral test for convergence

Theorem:
  Consider an $n_{0}$ and a non-negative, continuous function $f$ defined on the unbounded $[n_0,+\infty[$, on which it is monotone decreasing. Then 
  $\forall (p,q)\in\mathbb{N}^{2}$ such that $n_o\leq p <q$:
  $${\displaystyle \int _{p+1}^{q+1 }f(x)\,dx\leq \sum _{k=p+1}^{q }f(k)\leq \int _{p}^{q }f(x)\,dx}$$


i don't understand why they took the bounded from $1$ to $n+1$ instead of 
$n \geq 2,\ 1=n_0=p,\ q=n$
$$\displaystyle \int_{1+1}^{n+1}\dfrac{1}{x}dx\leq \sum_{k=1+1}^{n}\dfrac{1}{k}\leq \int_{1}^{n}\dfrac{1}{x}dx$$

The reason for the $n+1$ in $\displaystyle \int_1^{n+1} \frac{1}{x}\, dx \le \sum_{k=1}^n \frac{1}{k}$ can be seen from this diagram from Wikipedia

so $\displaystyle \int_1^{6} \frac{1}{x}\, dx \le \frac11+\frac12+\frac13+\frac14+\frac15$
In effect they took  $p=0, \, q=n$ for the left hand inequality, possibly because it is more natural, and then they took $p=1, q=n$ for the right hand inequality to get it to work.  For the left hand inequality, you only need finite values for $x=p+1$ and above 
