# $\underset{n\rightarrow +\infty }{\overset{}{\lim }} \ \prod\limits ^{2n}_{k=n}\sqrt[k]{k} =+\infty$

Show that:

$$\underset{n\rightarrow +\infty }{\overset{}{\lim }} \ \prod\limits ^{2n}_{k=n}\sqrt[k]{k} =+\infty$$

I thought to take:

$$e^{\sum\limits ^{2n}_{k=n}\dfrac{\ln k}{k}}$$

and now maybe Stolz-Cesaro?

This exercise is on first chapters of calculus textbook then I guess it should be possible to solve it without integral.

• So you want to show the sum in the exponent diverges. To do that, compare it with the integral $\int_n^{n+1} (\ln x)/x dx$. – user58955 Mar 7 at 15:09
• Thank you, this exercise is on first chapters of calculus textbook then I guess it should be possible to solve it without integral. – asv Mar 7 at 15:13

$$\sum\limits ^{2n}_{k=n}\dfrac{\ln k}{k} \geq (\ln n)(\sum\limits ^{2n}_{k=n}\dfrac{1}{k}) \geq (\ln n)(\sum\limits ^{2n}_{k=n}\dfrac{1}{2n}) \geq \frac{ \ln n}{2}$$ which goes to $$\infty$$
First note: $$x therefore if we can find an lower bound for the exponent and show that that diverges then we are done.
Note: $$\frac{1}{2} \ln(2n) = n \frac{\ln(2 n)}{2n} < \sum\limits_{k=n}^{2n}\frac{\ln(k)}{k}$$ therefore we can say
$$\sqrt{2n} = \exp\left(\frac{1}{2}\ln(2n)\right) < \exp\left(\sum\limits_{k=n}^{2n}\frac{\ln(k)}{k}\right)$$
Now, $$\sqrt{2n} \to +\infty$$ as $$n \to \infty$$ so we see that the exponential you asked also diverges.