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?

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.