Answer
Converges to $\ln 2$.
Work Step by Step
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists.
Consider $a_n=n(2^{1/n}-1)$
Apply limits to both sides.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}n(2^{1/n}-1)$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}n2^{1/n}-\lim\limits_{n \to \infty}(1)=\lim\limits_{n \to \infty}\dfrac{2^{1/n}-1}{1/n}$
Since, we can see that the limit has the form of $\frac{0}{0}$, so take the help of L-Hospital's rule.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{2^{1/n} \ln 2}{1}$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{2^{1/n} \ln 2}{1}=\ln 2$
Therefore, the sequence converges to $\ln 2$.
