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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.

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  • $\begingroup$ So you want to show the sum in the exponent diverges. To do that, compare it with the integral $\int_n^{n+1} (\ln x)/x dx$. $\endgroup$ – user58955 Mar 7 at 15:09
  • $\begingroup$ Thank you, this exercise is on first chapters of calculus textbook then I guess it should be possible to solve it without integral. $\endgroup$ – asv Mar 7 at 15:13
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$\sum\limits ^{2n}_{k=n}\dfrac{\ln k}{k} \geq (\ln n)(\sum\limits ^{2n}_{k=n}\dfrac{1}{k}) \geq (\ln n)(\sum\limits ^{2n}_{k=n}\dfrac{1}{2n}) \geq \frac{ \ln n}{2}$ which goes to $\infty$

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First note: $x <y \implies \exp(x) < \exp(y)$ 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.

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