TITLE: What is $\lim_{n\to\infty} n(2^{1/n} -1)$
QUESTION [2 upvotes]: How would you find the value of $\lim_{n\to\infty} n(2^{1/n} -1)$?
I tried taking the logarithm on both sides and it does not appear to be working.

REPLY [0 votes]: Let $x>1$. From $$\log(x)=\int_1^x\frac{dt}{t}$$ it follows that $$1-\frac{1}{x}<\log(x)<x-1$$ and $$\log(x)=n\log(x^{1/n}).$$ Then $$n(1-x^{-1/n})<\log(x)<n(x^{1/n}-1)$$ and so $$x^{-1/n} < \frac{\log(x)}{n(x^{1/n}-1)} < 1.$$ Now let $n\to\infty$.
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