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 
$n \geq 2,\ 1=n_0=p,\ q=n$
$$\displaystyle \int_{1+1}^{n+1}\dfrac{1}{x}dx\leq \sum_{k=1+1}^{n}\dfrac{1}{k}\leq \int_{1}^{n}\dfrac{1}{x}dx$$
 A: The value of $p$ is different for each of the inequalities.


*

*The inequality $\displaystyle \int_1^{n+1} \frac{1}{x}\, dx \le \sum_{k=1}^n \frac{1}{k}$ comes from taking $p=0$ and $q=n$ in the theorem.

*The inequality $\displaystyle \sum_{k=2}^n \frac{1}{k} \le \int_1^n \frac{1}{x}, dx$ comes from taking $p=1$ and $q=n$ in the proposition. (It should also be noted that there's a typo in your notes: the $\int$ was erroneously replaced by a $\sum$ in the second bit.


The second of these points yields
$$\sum_{k=1}^n \frac{1}{k} = 1 + \sum_{k=2}^n \frac{1}{k} \le 1 + \int_1^n \frac{1}{x}\, dx = 1 + \ln(n)$$
You could have done it the way that you suggest, but that would have yielded
$$\ln(n+1)-\ln 2 \le \sum_{k=2}^n \frac{1}{k} \le \ln(n)$$
and hence
$$\ln(n+1)-\ln(2)+1 \le \sum_{k=1}^n \frac{1}{k} \le \ln(n)+1$$
and you ultimately end up with the same result, since
$$\ln(n+1)-\ln(2)+1 \sim \ln(n) \text{ and } \ln(n)+1 \sim \ln(n) \text{ as } n \to \infty$$
A: 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 
A: 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.
