limit of factorial divided by sum of factorials

Question

I'd like to determine the limit as $n\rightarrow \infty$ following function:$$f\left(n\right)=\frac{n!}{\sum_{k=0}^{n}k!}.$$The following calculations suggest the answer is 1:
$$ \begin{align*} f\left(10\right)&\approx0.898682 \\\\ f\left(30\right)&\approx0.966627 \\\\ f\left(70\right)&\approx0.985711. \end{align*} $$ But this isn't very satisfying. Any input on how I might proceed in finding this limit would be appreciated.

Answer 1

If you write $$\sum_{k=0}^n k!=n!\Bigg[\frac 11 +\frac 1 {n}+ \frac 1 {n(n-1)}+ \frac 1 {n(n-1)(n-2)}+\frac 1 {n(n-1)(n-2)(n-3)}+\cdots \Bigg]$$ Expanding as series for large values of $n$ $$\sum_{k=0}^n k!=n!\Bigg[1+\sum_{k=1}^\infty \frac {B_k}{n^k}\Bigg]$$where appear Bell numbers. So, using long division $$f_n=\frac{n!}{\sum_{k=0}^{n}k!}=1-\frac 1n-\frac 1{n^3}-\frac 2{n^4}+O\left(\frac{1} {n^5}\right)$$

Using the above truncated expansion for $n=10$ as you did $$f_{10}=\frac{2247}{2500}=0.898800$$ to be compared to the exact value $$f_{10}=\frac{1814400}{2018957}=0.898682$$corresponding to a relative error of $0.013$%.