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Compute $$ \lim\limits_{n\to\infty}\left(\sqrt[n]{\log\left|1+\left(\dfrac{1}{n\cdot\log\left(n\right)}\right)^k\right|}\right). $$ What I have: $$ \forall\ x\geq 0\ :\ x- \frac{x^2}{2}\leq \log(1+x)\leq x. $$ Apply to get that the limit equals $1$ for any real number $k$.

Is this correct? Are there any other proofs?

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    $\begingroup$ You are using both notations $\log$ and $\ln$. I assume that they are the same? $\endgroup$ – Sammy Black Nov 6 '13 at 16:48
  • $\begingroup$ @AntonioVargas I am using the Squeeze Theorem. $\endgroup$ – Ahaan S. Rungta Nov 6 '13 at 17:09
  • $\begingroup$ @SammyBlack Yes, thanks! Edited. $\endgroup$ – Ahaan S. Rungta Nov 6 '13 at 17:09
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Yes it works, here's another proof using a little more sofisticate tool (in this case unnecessary, but sometimes more useful).

By Stolz-Cesaro if $ (x_n) $ is a positive sequence and $$ \lim_n \dfrac{x_{n+1}}{x_n} = l $$ then $$ \lim_n \sqrt[n]{x_n} = l.$$

Taking as $(x_n)$ the sequence you defined, an easy calculation shows that $$ \dfrac{x_{n+1}}{x_n} \rightarrow 1,$$ therefore the thesis.

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