# How find this $\lim_{n\to\infty}\frac{1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}}{n}$

Find this limits $$\lim_{n\to\infty}\dfrac{1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}}{n}$$

I want use $$\sqrt[n]{i}=e^{\dfrac{\ln{i}}{n}}\approx 1+\dfrac{\ln{i}}{n},1\le i\le n$$ but$$\lim_{n\to\infty}\dfrac{\ln{i}}{n}$$

and other idea is $$n<1+\sqrt[n]{2}+\cdots+\sqrt[n]{n}<?$$

three idea: I want use Stolz therom,

and last found this three idea is not usefull solve this limits

• Hint: $i^{1/n} = e^{\log i/n}\approx 1+\frac{\log i}{n}$... – Steven Stadnicki Nov 10 '14 at 6:48

Hint: $$n=1 + 1 + \cdots +1 \leq 1+\sqrt[n]{2}+\cdots+\sqrt[n]{n} \leq \sqrt[n]{n} + \sqrt[n]{n} + \cdots \sqrt[n]{n}=n \cdot \sqrt[n]{n}$$, and $$\text{lim}_{n\to\infty}\sqrt[n]{n}=1$$.
$$n<1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}<\int_0^{n+1}x^{\frac{1}{n}}dx=\frac{n}{n+1}(n+1)^{1+\frac{1}{n}}$$ $$1<\dfrac{1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}}{n}<(n+1)^{\frac{1}{n}}$$ $$\lim_{n\to\infty}\dfrac{1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}}{n}=1$$