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I am looking for a sequence $u_n$ of positive numbers such that $$\frac{u_{n+1}-u_{n}+1}{u_n}\sim \frac{c}{n\ln(n)},$$ as $n\to \infty$, for some $c>0$. I have a feeling that there is no such sequence. Here, $\sim$ means that $$\lim\frac{u_{n+1}-u_{n}+1}{u_n} \frac{n\ln(n)}{c}=1.$$

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What about $$ u_n=(1+\frac{c}{n})^{\frac{1}{\ln(n)}}$$ It is straightforwardly shown that $$u_n=1+\frac{c}{n\ln(n)} +O\left(\frac{1}{n^2\ln(n)}\right)$$ and $$\frac{u_{n+1}-u_{n}+1}{u_n}=\frac{c}{n\ln(n)} +O\left(\frac{1}{n^2\ln(n)}\right) $$

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  • $\begingroup$ Thank you Paul Enta. However the sequence $u_n$ that you have indicated goes to $1$ with $n\to\infty$. This seems to contradict you last equation... $\endgroup$ – GoldSoundz Jun 21 '17 at 3:31
  • $\begingroup$ I am sorry, I was not careful enough. $\endgroup$ – Paul Enta Jun 21 '17 at 10:49

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