2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Hint: $i^{1/n} = e^{\log i/n}\approx 1+\frac{\log i}{n}$... $\endgroup$ – Steven Stadnicki Nov 10 '14 at 6:48
4
$\begingroup$

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 $\text{lim}_{n\to\infty}\sqrt[n]{n}=1$.

You also might have a look at Cesàro's Theorem on averages of limits.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

$$n<1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}<\int_0^{n+1}x^{\frac{1}{n}}dx=\frac{n}{n+1}(n+1)^{1+\frac{1}{n}}$$ $$1<\dfrac{1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}}{n}<(n+1)^{\frac{1}{n}}$$ $$\lim_{n\to\infty}\dfrac{1+\sqrt[n]{2}+\sqrt[n]{3}+\cdots+\sqrt[n]{n}}{n}=1$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.