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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?

Iorpim
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    It is well known that if $\lim a_{n} = a$, then $\lim (a_{1}+\cdots + a_{n})/n = a$. – Seewoo Lee Sep 18 '18 at 00:18
  • There are also several issues in your post. You should have $\sum_{i = 1}^n 2^{1/i}$ rather than $2^{1/n}$, and the assumption that $\sum 2^{1/i}$ converges to $a \in \mathbb{R}$ is not correct. –  Sep 18 '18 at 00:21

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