$$\lim_{n\to\infty}\frac1{n^{n+1}}\sum_{k=0}^n\left(k\binom nk\left(\sum_{i=0}^k(-1)^{k+i}\binom ki i^n\right)\right)$$
I know this is the right answer because I approximated this for $n=1000$ and the app said answer is close but not near enough to accept. It needs 6 digits of precesion or exact answer. It's about $0.632$. How do I solve this summation.
Here's the exact question for anyone wondering
You are given a collection of $N$ distinct objects from which you draw a random sample of size $N$ with replacement — that means objects are allowed to be sampled multiple times. Generating a sample in this way is related to the statistical technique called bootstrapping.
We're interested in the expected fraction of the original collection that ends up in the sample (i.e. the probability that a given object in the original collection appears at least once in the sample). What does this probability converge to as $N$ goes to infinity?
The inner sum is associated to the application of the $k$-th power of the forward difference operator to the polynomial $z^n$. In particular the inner sum is $k!{n\brace k}$, with ${n\brace k}$ being a Stirling number of the second kind. So we have to deal with $$ \frac{1}{n^{n+1}}\sum_{k=0}^{n}k\binom{n}{k}k!{n\brace k}\tag{1}$$ where $$ \sum_{k=0}^{n}\binom{x}{n}k!{n\brace k} = x^n \tag{2}$$ and $k\binom{n}{k} = n\binom{n-1}{k-1}$ for any $k,n\geq 1$. Thus the written limit is $$ 1-\lim_{n\to +\infty}\left(1-\frac{1}{n}\right)^n = 1-\frac{1}{e}.$$