How would one prove that
$\lim_{n\rightarrow\infty} \log(n) \left( 1 - \left(\frac{n}{n-1}\right)^{\log(n)+1}\right) = 0 $ ?
I can see from Mathemtica that the limit is zero.
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We have using Taylor series:
$$1-\left(\frac{n}{n-1}\right)^{\log n+1}=1-\exp\left((\log n+1)\log\left(\frac{n}{n-1}\right)\right)\\=1-\exp\left(\underbrace{(\log n+1)\left(\frac1n+o\left(\frac1n\right)\right)}_{=:u_n\xrightarrow{n\to\infty}\;0}\right)=1-1-u_n+o(u_n)\sim_\infty u_n$$ and we know that
$$u_n\log n\xrightarrow{n\to\infty}0$$ since by simple use of L'Hôpital's rule we have $$\frac{\log^2 n}{n}\xrightarrow{n\to\infty}0$$
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The same arguments appeared at the same time. Sorry I can only accept one. – user39531 Aug 05 '14 at 16:23
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$$ \dfrac{n}{n-1} = 1 + \dfrac{1}{n} + O(1/n^2)$$ $$ \log\left( \dfrac{n}{n-1}\right) = \log\left( 1 + \dfrac{1}{n} + O(1/n^2)\right) = \dfrac{1}{n} + O(1/n^2)$$ $$ \eqalign{\left( \dfrac{n}{n-1} \right)^{\log(n)+1} &= \exp\left(\log\left( \dfrac{n}{n-1} \right) (\log(n) + 1)\right)\cr &= \exp\left(\dfrac{\log(n)}{n} + O(1/n)\right)\cr &= 1 + \dfrac{\log(n)}{n} + O(1/n)} $$ $$ \log(n) \left(1 - \left( \dfrac{n}{n-1} \right)^{\log(n)+1}\right) = -\dfrac{\log(n)^2}{n} + O(\log(n)/n) \to 0 $$
Robert Israel
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It is easy to see that $$1\lt\left(\frac{n}{n-1}\right)^{\ln(n)+1}\lt e^{\frac{\ln(n)+1}{n-1}},$$ on the other hand we have $$\lim_{n\rightarrow \infty}\ln(n)\left(e^{\frac{\ln(n)+1}{n-1}}-1\right)=\lim_{n\rightarrow\infty}\frac{\ln(n)(\ln(n)+1)}{n-1}\frac{e^{\frac{\ln(n)+1}{n-1}}-1}{\frac{\ln(n)+1}{n-1}}=0,$$ and we are done.
pointer
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