Prove that the sequence converges Prove that the sequence converges.

For each positive integer $n$, let
  $$y_n = 1 + \frac12 + \frac13 + \cdots + \frac1n - \int_1^n \frac{dx}x.$$

 A: 
The blue-curve takes the value $\frac1{k}$ over an interval $[k,k+1)$
The red-curve is given by $f(x) = \frac1{x}$ where $x \in [1,\infty)$
The green-curve takes the value $\frac1{k+1}$ over an interval $[k,k+1)$
The area under the blue-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k}$
The area under the red-curve is given by the integral $\displaystyle \int_{1}^{n+1} \frac{dx}{x}$
The area under the green-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k+1}$
Can you now formalize this argument?
A: I don't see quite this approach at this or the older question. Along with the sequence $y_n$ as defined, also define a second sequence
$$  z_n = y_n - \frac{1}{n}.$$
Note $y_1 = 1,$ while $z_1 = 0.$ 
So  we always have $y_n > z_n.$  Prove  that the $y_n$ are decreasing in $n,$ while the $z_n$ are increasing in $n.$ So, all the $y_i$ are greater than all the $z_j.$ But  $y_n - z_n =  \frac{1}{n}$ becomes arbitrarily small.
I did a little program on a calculator, I get $y_{22} < 0.6,$ while $z_7 > 0.5.$ 
As I looked at this again, the only calculus part is the necessary proof that
$$ \frac{1}{n+1} <  \int_n^{n+1} \frac{dx}{x}   <       \frac{1}{n}   $$ 
which is easy enough.
