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Let $(x_n)$ be a sequence such that $\lim_{n\to\infty}\frac{1}{n}\log(x_n)=L$ with $L<\infty$. My (maybe silly) question is, whether $$ \lim_{n\to\infty}\frac{1}{n}\log(x_n+1)=L? $$

Sad, but I do not know how to prove/ disprove that therefore please give me some help.

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If $\lim_{n\to\infty}\frac1n \log(x_n) = L$, then $\log(x_n)\sim Ln$ as $n\to\infty$. Therefore $x_n\to\infty$ as $n\to\infty$. Hence $${x_n\over 1 + x_n}\to 1$$ as $n\to\infty$, provided $L > 0$.

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  • $\begingroup$ Does this mean that $x_n\sim 1+x_n$, thus $\log(x_n)\sim \log(x_n+1)$, so that my question is to answer with yes in case $L>0$? $\endgroup$ – math12 Apr 30 '15 at 18:59
  • $\begingroup$ @ncmathsadist I took the liberty of inserting the previously missing term $1/n$ from the first limit expression. I hope that you don't mind. And +1 $\endgroup$ – Mark Viola Apr 30 '15 at 19:28
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Take $x_n=e^{-n}$, then $\lim_{n\to\infty}\frac{1}{n}\log(x_n)=-1$ but $\lim_{n\to\infty}\frac{1}{n}\log(x_n+1)=0$.

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  • $\begingroup$ This happens whenever $L < 0$. $\endgroup$ – ncmathsadist Apr 30 '15 at 18:48
  • $\begingroup$ What, if $L=0$? $\endgroup$ – math12 Apr 30 '15 at 19:09

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