TITLE: Limit with log in exponent
QUESTION [2 upvotes]: 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.

REPLY [0 votes]: 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.
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