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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 $\mathbb{R}$?

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    Basically integrate the whole expression (1) over $[1,n]$ as suggested, but divide through by $n$ first. The sum on the left then integrates to $-u_n$. The first term in the integral on the RHS, $\int_1^n \frac{1}{x}dx$, becomes $\ln(n)$, and the second term is the integral you are after (call it $I$).

    This gives you $-u_n = \ln(n) - I$, i.e. $v_n = I$.

     – fGDu94 Dec 04 '19 at 10:27
  • Hello @GeorgeDewhirst thanks , could you please write a complete answer I'm still stuck for the sum I don't find $-u_n$ –  Dec 04 '19 at 15:34
  • Still no clue :\ –  Dec 05 '19 at 16:46
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    oh sorry I'll get right to it.

    the integral of $(1)$ from 1 to n gives:

    LHS = $-n\sum_{k=0}^n (1-\frac{x}{n})^k = - n u_n$. RHS = $n(\ln(n) - n v_n)$

     – fGDu94 Dec 05 '19 at 16:53
  • Ouf finally I get that $-nu_n$ thanks a lot @GeorgeDewhirst –  Dec 06 '19 at 01:01
  • Sorry to bother you again @GeorgeDewhirst but how can we conclude with the result that $(v_{n})_{n\geq 1}$ is convergent ? –  Dec 06 '19 at 03:28
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    $v_n$ is monotone increasing and is bounded above by $\int_1^{\infty}\frac{e^{-x}}{x}dx<\infty$, thus it is convergent. – fGDu94 Dec 06 '19 at 03:42

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