2
$\begingroup$

Help with Limit: $$\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}$$

EDIT: Posted a similar question just yesterday but bad formating from my part dropped a term, and lead me to strange places. Honestly, I am just plain stuck, I have been hitting my head against it for $2$ days straight. I know the solution should be $2^{-1/2}$ but...

Help would be appreciated. If anyone has a Wolfram Alpha PRO account I am sure that thing churns out the solution as it is something quite basic I just don't see it and its driving me crazy.

$\endgroup$
  • 1
    $\begingroup$ Wanted to delete this one, but the community was way too fast. $\endgroup$ – Katpton Liamfuppinshire Nov 8 '16 at 6:59
4
$\begingroup$

Hint: Use Stirling's approximation $$n!\approx n^n e^{-n}\sqrt{2\pi n}$$ for large $n$.

$\endgroup$
3
$\begingroup$

Using Stirling's approximation, we have that \begin{align} \frac{(2n)!e^nn^n}{n!(2n)^{2n}} \approx \sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n} \frac{e^nn^n}{(2n)^{2n}} \frac{1}{\sqrt{2\pi n}}\left( \frac{e}{n}\right)^n = \sqrt{2}. \end{align}

$\endgroup$
2
$\begingroup$

Hint. Use Stirling approximation: $$n!\sim\sqrt{2\pi n} \cdot\frac{n^n}{e^n}.$$ which implies that $$(2n)!\sim2\sqrt{\pi n} \cdot\frac{(2n)^{2n}}{e^{2n}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.