Snake Oil

Actually, this is more of a combinatorics problem.Let us evaluate f(n).Firstly, take (-1)^n out of the summation, so that our summand becomes (-1)^r times (r^n)(nCr).To evaluate this, I consider the following scenario: "Suppose there are n students and n prizes.We want to distribute the prizes among the students such that every student gets atleast one prize."Clearly every student gets exactly one prize, hence the total number of distributions are n! .Now, let us solve this scenario using a different method.Suppose that I choose r students (1<r <n) and distribute the prizes among these r students.It can be done in (nCr)(r^n) ways.But this includes those cases also in which some students out of the r chosen students do not get any prize.Hence, by principle of inclusion and exclusion, the number of ways in which n prizes can be distributed among n students, such that every student gets atleast one prize will be : (nCn)(n^n) - (nC(n-1))((n-1)^n) + (nC(n-2))((n-2)^n) - .......... which is nothing but the summation asked in the question in reverse order.Hence we get that f(n)=n! and the limit becomes 1.

Define $\displaystyle \Delta (f(x)) = f(x+1) - f(x)$ . Also, $\text{E} (f(x)) = f(x+1)$ . Clearly, $\Delta \equiv \text{E} - 1$ .

Now,

$\displaystyle \Delta^{n} (f(x)) = (\text{E} - 1)^{n} f(x)$

$\displaystyle = (-1)^n \sum_{r=0}^{n} (-1)^r \dbinom{n}{r} \text{E}^{r} (f(x))$

$\displaystyle = (-1)^n \sum_{r=0}^{n} (-1)^r \dbinom{n}{r} f(x+r) \tag{*}$

where $\displaystyle \Delta^{n} (f(x))$ represents $n$ compositions of $\Delta$ over $f$ .

Note that,

$\displaystyle \Delta^{n} (x^m) = 0 \ \forall \ n > m$ , so for $f(x) = x^n$ , we have,

$\displaystyle \lambda = \Delta^{n} x^n = \Delta^{n-1} (\Delta (x^n)) = \Delta^{n-1} \left( \dbinom{n}{1} x^{n-1} + \ (\text{lower degree terms}) \right)$

$\displaystyle = \Delta^{n-2} \left( \dbinom{n}{1} \dbinom{n-1}{1} x^{n-2} + \ (\text{lower degree terms}) \right)$

Continuing like this, we have,

\displaystyle \lambda = n! \tag{#}

From $(*)$ and (#) , we have,

$\displaystyle \sum_{r=0}^{n} (-1)^{n-r} \dbinom{n}{r} (x+r)^n = n!$

Putting $x=0$ , we have,

$\displaystyle f(n) = \sum_{r=0}^{n} (-1)^{n-r} \dbinom{n}{r} r^n = n!$

$\displaystyle \implies \lim_{n \to \infty} \dfrac{f(n)}{n!} = \boxed{1}$