# Compute $\lim\limits_{n\to\infty}\sqrt[n]{\log\left|1+\left(\frac{1}{n\cdot\log n}\right)^k\right|}$.

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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You are using both notations $\log$ and $\ln$. I assume that they are the same? –  Sammy Black Nov 6 '13 at 16:48
@AntonioVargas I am using the Squeeze Theorem. –  Ahaan S. Rungta Nov 6 '13 at 17:09
@SammyBlack Yes, thanks! Edited. –  Ahaan S. Rungta Nov 6 '13 at 17:09

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.