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 
$n \geq 2,\ 1=n_0=p,\ q=n$
$$\displaystyle \int_{1+1}^{n+1}\dfrac{1}{x}dx\leq \sum_{k=1+1}^{n}\dfrac{1}{k}\leq \int_{1}^{n}\dfrac{1}{x}dx$$

You do not really need the integral test to prove it. Now that Landau's notation is your forte, you may simpy notice that
$$ \text{arctanh}\frac{1}{k}=\frac{1}{2}\log\left(\frac{1+\frac{1}{k}}{1-\frac{1}{k}}\right)=\sum_{n\geq 0}\frac{1}{(2n+1)\,k^{2n+1}}=\frac{1}{k}+O\left(\frac{1}{k^3}\right)\tag{1} $$
implies:
$$ H_n = \sum_{k=1}^{n}\frac{1}{k} = O(1)+\frac{1}{2}\sum_{k=2}^{n}\left[\log(k+1)-\log(k-1)\right] \tag{2} $$
where the last sum is a telescopic sum.
