Evaluating $ \lim_{n \to \infty} {\sqrt[n]{\frac{(2n - 1)!}{n!}}} $

Question

I have such limit to solve: $$ \lim_{n \to \infty} {\sqrt[n]{\frac{(2n - 1)!}{n!}}} $$ I understand that with the n-root I should go to $e^{\ln n}$, but the real problem cause factorials. What can I do with them?

Answer 1

The key is that $(n!)^{1/n} \approx \dfrac{n}{e} $ and $n^{1/n}\to 1$

So

$\begin{array}\\ \left(\dfrac{(2n - 1)!}{n!}\right)^{1/n} &=\dfrac{(2n - 1)!^{1/n}}{n!^{1/n}}\\ &=\dfrac{(2n)!^{1/n}}{(2n)^{1/n}n!^{1/n}}\\ &\approx\dfrac{((2n)!^{1/(2n)})^2}{(2n)^{1/n}n!^{1/n}}\\ &\approx\dfrac{(2n/e)^2}{n/e}\\ &\approx \dfrac{4n}{e}\\ &\to \infty\\ \end{array} $

Answer 2

Hint: Use Stirling's approximation

\begin{align*} n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \end{align*}

Answer 3

$$\sqrt[n]{\frac{(2n-1)!}{n!}}=\sqrt[n]{\prod_{k=1}^{n-1} (n+k)}\ge \sqrt[n]{\prod_{k=1}^{n-1} n}= \sqrt[n]{n^{n-1}}=n(\sqrt[n]{n})^{-1}\to+\infty$$ because $\sqrt[n]{n}\to 1$

Answer 4

Consider the more general problem of $$\lim_{n \to \infty} {\sqrt[n]{\frac{(an +b)!}{n!}}}$$ Let $$A_n={\sqrt[n]{\frac{(an +b)!}{n!}}}\implies \log(A_n)=\frac 1n \left(\log[(an+b)!]-\log(n!]\right)$$ Using twice Stirling approximation and continuing with Taylor series for large $n$, then $$ \log(A_n)=a\log(a)+(a-1)\log \left(\frac{n}{e}\right)+\frac{\left(b+\frac{1}{2}\right) \log (a)+b \log (n)}{n}+O\left(\frac{1}{n^2}\right)$$ Notice that $b$ does not play any role in the limit.

Now, consider the three cases : $0<a<1$, $a=1$ and $a>1$