Prove that $v_n=\int_{1}^{n}{(1-\frac{x}{n})^{n}} \times \frac{1}{x} \, dx$

Question

It would be really helpful if anyone would help for this question

We have :

$$u_{n}=\sum_{k=1}^{n}{(1-\frac{1}{n})^{k}}\;\;\;,\;\;\;v_n=ln(n)\,-\,u_n$$

I proved $\,\forall\,n \ge1 \, ,x>0 $

$\;\;\;\:\:\;\:\:\:\;\;\;\:\:\:\;\;\;\:\:\:\;\;\;\:\:\:1 + ( 1 - \frac{x}{n}) + ...... + ( 1 - \frac{x}{n})^{n-1} = \frac{n}{x}(1-(1-\frac{x}{n})^{n}) \;\;\;\:\:\:$(1) ( Geometric series)

Now how can we have to prove that :

$$v_n=\int_{1}^{n}{(1-\frac{x}{n})^{n}} \times \frac{1}{x} \, dx$$

using the Lebesgue integral on the interval $[1,n]$ for each part of (1) was a hint that we had but I don't know we can proceed.

Thanks for your help

Edit : Can we conclude that $(v_{n})_{n\geq 1}$ is convergent on $\mathbb{R}$?