There is the following limit, I would like to calculate:
$$\lim_{n\rightarrow\infty} \frac{2n}{(n!)^{1/n}}$$
After the substituion with Stirling's approximation I have got a relatively complex formula and I can't seem to solve it
On
By Stirling, $(n!)^{1/n} \sim \frac{n}{e} $, since $n^{1/n} \to 1$ and $c^{1/n} \to 1$ for any $c > 0$.
On
Comment continuing from the Answer of @Victor: First, see the various versions of Stirling's approximations on the Internet, especially Wikipedia. Also, computation (in R) gives the following, where it was necessary to take logs in order to avoid overflow for larger values of n:
n = c(2:5, 10, 50, 100, 1000, seq(10000, 100000, by = 10000))
lnum = log(2*n)
lden = (1/n)*lfactorial(n)
lfrac = lnum - lden
frac = exp(lfrac)
cbind(n, lnum, lden, lfrac, frac)
n lnum lden lfrac frac
[1,] 2e+00 1.386294 0.3465736 1.039721 2.828427
[2,] 3e+00 1.791759 0.5972532 1.194506 3.301927
[3,] 4e+00 2.079442 0.7945135 1.284928 3.614408
[4,] 5e+00 2.302585 0.9574983 1.345087 3.838519
[5,] 1e+01 2.995732 1.5104413 1.485291 4.416250
[6,] 5e+01 4.605170 2.9695553 1.635615 5.132613
[7,] 1e+02 5.298317 3.6373938 1.660924 5.264171
[8,] 1e+03 7.600902 5.9121282 1.688774 5.412842
[9,] 1e+04 9.903488 8.2108928 1.692595 5.433561
[10,] 2e+04 10.596635 8.9037811 1.692854 5.434968
[11,] 3e+04 11.002100 9.3091551 1.692945 5.435463
[12,] 4e+04 11.289782 9.5967902 1.692992 5.435719
[13,] 5e+04 11.512925 9.8199049 1.693021 5.435876
[14,] 6e+04 11.695247 10.0022068 1.693040 5.435982
[15,] 7e+04 11.849398 10.1563433 1.693054 5.436059
[16,] 8e+04 11.982929 10.2898640 1.693065 5.436118
[17,] 9e+04 12.100712 10.4076385 1.693074 5.436164
[18,] 1e+05 12.206073 10.5129922 1.693080 5.436201
2*exp(1) # 2e
[1] 5.436564
$$n! \sim \sqrt{2 \pi n} \ \bigg(\frac{n}{e} \bigg)^n \implies n!^{\frac{1}{n}} =\sqrt{2 \pi n} ^{\frac{1}{n}} \ \frac{n}{e} $$
First factor, namely $\sqrt{2 \pi n} ^{\frac{1}{n}} $ obviously $\rightarrow 1$ as $n \rightarrow \infty$. You're consequently left only with:
$$ \frac{2n}{\frac{n}{e}}$$
I'm sure you can take it from here.