Limit $\lim_{n \to \infty}\frac{n!-\sin(2n)}{\sqrt{(n+1)!}-\ln(n^2+3n)}$

Question

My professor of Analysis has said that to simplify the factorial I can use Stirling's formula, but I don't know very well what it is or how to apply it.

$$\lim_{n \to \infty}\dfrac{n!-\sin(2n)}{\sqrt{(n+1)!}-\ln(n^2+3n)}$$

I have tried with Stolz but it doesn't work.

Answer 1

Stirling Formula gives an good approximation of how $n!$ behaves when $n \rightarrow +\infty$, it is given by $$ n! \underset{(+\infty)}{\sim}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$Indeed you can use it here, but careful, you cannot sum the $\sim$ relation.

However you can use that $$ \frac{n!-\sin\left(2n\right)}{n!}=1-\frac{\sin\left(2n\right)}{n!} \underset{n \rightarrow +\infty}{\rightarrow}1 $$ And $$ \frac{\sqrt{(n+1)!}-\ln\left(n^2+3n\right)}{\sqrt{(n+1)!}}=1-\frac{\ln\left(n^2+3n\right)}{\sqrt{(n+1)!}}\underset{n \rightarrow +\infty}{\rightarrow}1 $$ Hence

$$ \frac{n!-\sin\left(2n\right)}{\sqrt{(n+1)!}-\ln\left(n^2+3n\right)}\underset{(+\infty)}{\sim}\frac{n!}{\sqrt{(n+1)!}} $$

Answer 2

$$\lim_{n \to \infty}\dfrac{n!-\sin(2n)}{\sqrt{(n+1)!}-\ln(n^2+3n)}=\lim_{n \to \infty}\dfrac{\sqrt{\frac{n!}{n+1}}-\frac{\sin(2n)}{\sqrt{(n+1)!}}}{1-\frac{\ln(n^2+3n)}{\sqrt{(n+1)!}}}=\infty$$

Answer 3

The idea is that $\sin$ and $\ln$ are irrelevant in the computation. Thus, the limit dependens only on the ratio between the two factorial expressions, the one at the denominator being "almost" the square root of the other one. Formalizing:

$\begin{align}\lim_{n \to \infty} \frac{n!-\sin(2n)}{\sqrt{(n+1)!}-\ln(n²+3n)}\sim\\ \sim\lim_{n \to \infty} \frac{n!}{\sqrt{(n+1)}n!^{\frac{1}{2}}}=\\ =\lim_{n \to \infty} \sqrt{\frac{n!}{n+1}} \to +\infty\end{align}$

Answer 4

For this kind of limits the key point is to recognize that $n!$ and $\sqrt{(n+1)!}$ are "bigger" than all others terms, thus an efficent way to solve is to collect them

$$\dfrac{n!-\sin(2n)}{\sqrt{(n+1)!}-\ln(n^2+3n)}=\frac{n!}{\sqrt{(n+1)!}}\dfrac{1-\frac{\sin(2n)}{n!}}{1-\frac{\ln(n^2+3n)}{\sqrt{(n+1)!}}}\to+\infty\cdot\frac{1-0}{1-0}=+\infty$$

Answer 5

$ \dfrac{n!-\sin(2n)}{\sqrt{(n+1)!} - \log(n^2+3n)} \gt$

$\dfrac{n! -1}{\sqrt{(n+1)!}} \gt$

$\dfrac{n!}{\sqrt{(n+1)n!}} -\dfrac{\sqrt{n+1}}{\sqrt{(n+1)!}}=$

$\sqrt{n!}\dfrac{1}{\sqrt{(n+1)}} -\dfrac{1}{\sqrt{n!}} \gt$

$\sqrt{(n-2)!} - \dfrac{1}{\sqrt{n!}}.$

The limit $n \rightarrow \infty$ is ?

Used: $n(n-1) \gt n+1$, for $n\ge 3.$