Limit as $n$ goes to infinity of $n^22^n/n!$

Question

I have read in a book that $( n^2 2^n)$ is superior to n! this means that the limit below will at least be a constant and $n! = O( n^2 2^n )$, but l could not manage to find it, any ideas $$\lim_{n \to \infty} \frac{\ n^2 2^n}{\ n!}$$

Answer 1

Just transform everything into exponential like

$$\frac{e^{2\ln(n)}\cdot e^{n\ln(2)}}{e^{\ln(n!)}}$$

Now use the fact that $\ln(n!) \approx n\ln(n) - n$ and substitute using powers properties obtaining

$$e^{n\ln(2) + 2\ln(n) - n\ln(n) + n}$$

namely

$$e^{n\cdot[\ln(2) - \ln(n) + 1] + 2\ln(n)}$$

as $n\to\infty$ you can easily see that what remains is simply

$$e^{-n\ln(n)}$$ which goes to zero pretty fast.

Answer 2

Stirling's formula states that $$\lim_{n \rightarrow \infty} \frac{n!}{\sqrt{2\pi n} (\frac{n}{e})^n} = 1$$ so $$ \lim_{n \rightarrow \infty} \frac{n^22^n}{n!} = \lim_{n \rightarrow \infty} \frac{n^22^n}{\sqrt{2\pi n}(\frac{n}{e})^n} $$ and $$\lim_{n \rightarrow \infty} \frac{n^22^n}{\sqrt{2\pi n}(\frac{n}{e})^n} = \lim_{n \rightarrow \infty} \frac{(2e)^n}{n^{n-2}\sqrt{2\pi n}} = \lim_{n \rightarrow \infty} (2e)^2 (\frac{2e}{n})^{n-2} \frac{1}{\sqrt{2\pi n}}= 0$$ which disagrees with the book. WolframAlpha also says that the limit is $0$, so perhaps you misunderstood what the book said or the book contains an error.

Answer 3

Using logarithms makes this kind of problems simpler. Considering $$A=\frac{\ n^2 2^n}{\ n!}$$ $$\log(A)=2\log(n)+n\log(2)-\log(n!)$$ Now, Stirling approximation $$\log(n!)=\log(\sqrt{2\pi})+(n+\frac 12)\log(n)-n+\frac 1 {12n}+\cdots$$ So $$\log(A)=-\log(\sqrt{2\pi})+(1+\log(2))n-(n-\frac 32)\log(n)-\frac 1 {12n}+\cdots$$ from which you can conclude that $\log(A)\to -\infty$ and $A\to 0$.

By the way $7^2\times 2^7=6272$ while $7!=5040$ but $8^2\times 2^8=16384$ while $8!=40320$.

In the real domain $x^2 \times 2^x \gt x!$ if $x\lt 7.20737$.