Why is $\lim_{n\rightarrow \infty } \frac{\sum_{i=1}^{n} 2^{1/n}}{n}$ equal to 1?

Question

As the title states, why is this? The original sequence is $b_n=\frac{2+\sqrt[]{2}+\sqrt[3]{2}+\cdots +\sqrt[n]{2}}{n}$, however: $$b_n=\frac{2+\sqrt[]{2}+\sqrt[3]{2}+\cdots+ \sqrt[n]{2}}{n}=\frac{\sum_{1}^{n} 2^{1/i}}{n}\Rightarrow \lim_{n\rightarrow \infty } b_n=\lim_{n\rightarrow \infty } \frac{\sum_{1}^{n} 2^{1/i}}{n}$$ I would assume $\sum_{1}^{n} 2^{1/i}$ to converge to an $a$, as $lim_{n\rightarrow \infty }\sqrt[n]{2}=1$, so that $\lim_{n\rightarrow \infty } \frac{a}{n}$ would go to 0, however, the result is 1, why is that? And, how could I properly evaluate this expression?