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Let the sequence $(a_n)_{n \geq 1}$ be defined by $a_1 >0 $, and the recurrent relation $$a_n = 2a_{n-1}*\frac{(n-1)^{(n-1)}}{(n-1)!}$$

Then, what is $\lim_{n\to\infty} \frac {{a_{n+1}}^{1/{n+1}}}{{a_n}^{1/n}} $?

So far, I've managed to prove that $\lim_{n\to\infty}{(\frac{a_{n+1}}{a_n})}^{1/n}=e,$ by using the Stirling limit. Therefore, it should suffice to calculate $\lim_{n\to\infty} {{a_{n+1}}^{\frac{-1}{n(n+1)}}}. $

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    $\begingroup$ We only have a power of the form $x^x$, I would omit the "tetration" in the title, although we have actually a simple case of it. $\endgroup$ – Peter Feb 23 '20 at 10:28
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A linear recurrence of the first order, homogeneous. Easy to see that the solution is:

$$ a_n = a_1 2^ n \prod_{1 \le k \le n} \frac{k^k}{k!} $$

Compute the limit from here. Much of the mess should simplify away.

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  • $\begingroup$ Hello, I've took notice of this fact already - however much of the mess did not simplify away... because of the different exponents of $a_{n+1}$ and $a_n$ in the limit. $\endgroup$ – Parallelism Alert Feb 23 '20 at 12:06
  • $\begingroup$ What is $\lim_{n \to \infty}(n + 1)/n$? $\endgroup$ – vonbrand Feb 23 '20 at 14:05
  • $\begingroup$ I don't see where you are going with this hint.... $\endgroup$ – Parallelism Alert Feb 23 '20 at 16:25

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