Prove that the sequence converges

Prove that the sequence converges.

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

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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?

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.