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?
Answer
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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