mjhjmtu

$$(x_n)_ngeq 2 x_n=sqrt[n]1+sum_k=2^n(k-1)(k-1)! $$

$$lim_nrightarrow infty fracx_nn=?$$

I'm trying to solve the sum I but I don't know how to rewrite it.Can you give me some hints?

edit:

The sum $sum_k=2^n(k-1)(k-1)!$ should be $(n! - 1)$ so $x_n=sqrt[n]n!$.The limit would be $lim_nrightarrow infty sqrt[n]fracn!n^n$ .If I use cauchy d'alembert criterion and make some simplifications I would get $e^-1$ (the right answer)My problem is that sum.How I obtain $n! - 1$ ?

Very roughly, the sum will be dominated by the last term. You can then apply Stiriling's approximation to the last term
$$(n-1)(n-1)!approx frac (n-1)^ne^n-1sqrt 2pi (n-1)$$
The $n^th$ root of this goes to $frac n-1e$ so the limit goes to $frac 1e$
You need to justify the approximations made, particularly the one about considering only the last term of the sum.

Well, using Stirling's formula $m!=(m/e)^msqrt2pi m(1+o(1))$ as $mtoinfty$, we find that the radicand for large $k$ is $((n-1)/e)^n-1 A_n$, where $1le A_nle Cn^5/2$ with some positive constant $C$ independent of $n$. Hence $lim_ntoinftysqrt[n]A_n=1$, and we obtain
$$
lim_ntoinfty fracx_nn=lim_ntoinftyfrac(n-1)^(n-1)/nne^(n-1)/n=frac1e.
$$

Note that $(k-1)(k-1)!$ is very nearly $k(k-1)!$, which is $k!$. We'll use that:
$$
sum_k=2^n (k-1)(k-1)! + sum_k=2^n (k-1)! = sum_k=2^n (k-1+1)(k-1)! = sum_k=2^n k!
$$
Then (careful with the limits):
$$
sum_k=2^n (k-1)(k-1)! = sum_k=2^n k! - sum_k=2^n (k-1)! = sum_k=2^n k! - sum_k=1^n-1 k! = n! - 1! tag*
$$
because almost all the terms cancel out.

Edit: neater train of thought, but technically the same:

Observe that $(k-1)(k-1)! = k! - (k-1)!$, and then follow with $(*)$.