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Problem Let $(u_n)$ the real sequence defined by $u_0 \in \mathbb{R}$ and $u_{n+1}=u_{n}+e^{-u_n}$ for all $n \in \mathbb{N}$. Show that $u_n=\ln n + \frac{\ln n}{2n}+o\left(\dfrac{\ln n}{n}\right)$

My attempt: First, I will prove that $(u_n)$ tends to $+\infty$. Indeed,

  1. Obviously, $(u_n)$ is an increasing sequence.
  2. Suppose that $(u_n)$ is bounded. Then $\lim u_n = L$. Then, we have $L=L+e^{L}$. That implies $e^{L}=0 (contradiction)$. Finally, we have $\lim u_n = +\infty$.

I intends to define $(v_n)$ such that $v_n=\exp(-u_n)$. But I'm not sure about my idea. Thank you for your help

zwim
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Pipnap
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  • Related: https://math.stackexchange.com/q/58866/42969 – Martin R Oct 07 '23 at 12:15
  • Here is a solution – NN2 Oct 07 '23 at 12:16
  • @NN2 thank you. But do we have any different approaches such as compute the limitation in https://math.stackexchange.com/questions/58866/lim-limits-n-to-infty-bigx-n-lnn1-big-for-x-n1-x-ne-x-n – Pipnap Oct 07 '23 at 13:22
  • @MartinR Yeah I tried to read that topic. But the computation in my case will be more complicated. I guess – Pipnap Oct 07 '23 at 13:23
  • @Pipnap The solution in the link proposed by MartinR is like mine (we both use the Stolz-Cesàro theorem. Even I think my solution for the first term, equation $(2)$, is simpler). And my solution is longer just because we need to develop the second asymptotic term $\ln(n) /(2n)$ $$$$ Besides, I think you can define $v_n = e^{-u_n}$ (I defined $v_n = e^{u_n}$) and obtain the same result as mine – NN2 Oct 07 '23 at 13:27
  • @NN2 Can you explain the line that you used Stolz - Cesaro. I saw that on wiki. But maybe you apply a version for O notation. I am not really familiar with that. – Pipnap Oct 07 '23 at 13:39
  • @Pipnap it's just a version with big O: If $\frac{x_{n+1}-x_n}{y_{n+1}-y_n} = l + \mathcal{O}(v_n)$ (with $v_n = \mathcal{o}(1)$) then $\frac{x_n}{y_n} = l + \mathcal{O}(v_n)$. You can forget the term $\mathcal{O}(v_n)$ if you think it's complicate. – NN2 Oct 07 '23 at 13:42
  • @NN2 thanks you. It tried to compute the limitation $\lim \frac{x_n - \ln n - \frac{\ln n}{2n}}{\frac{\ln n}{n}}$ but I can't hahaha – Pipnap Oct 07 '23 at 13:46
  • @NN2 Hi, can you explain why you can write $\underbrace{\mathcal{O}\left(\sum_{i=1}^{n-1} \frac{1}{i^2}\right)}_{\text {converge, }=o(\ln (n))}$ – Pipnap Oct 07 '23 at 14:29
  • @Pipnap because $\sum_{}i=1^{+\infty} 1/i^2 = \frac{\pi^2}{6} = \text{const} =\mathcal{o} (\ln(n))$ – NN2 Oct 07 '23 at 14:49
  • @NN2 we can replace ln (n) with any sequence which is >> $\frac{\pi ^2}{6} ? Do you have any lecture note related to this topic?. Anw, thank you a lot for your kind support. – Pipnap Oct 07 '23 at 14:56
  • Of course you can replace it with any sequence that is asymptotically greater than a constant. I don't have any lecture note but you can find many on the internet. – NN2 Oct 07 '23 at 15:23
  • Yes. Thank you a lot @leonbloy – Pipnap Oct 08 '23 at 22:38

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