Numerically evaluate a series involving Stirling numbers of the second kind How can I numerically evaluate the following for $n$ above one million:
$$ \sum_{k>=n} \sum_{i=0}^{n-1} (-1)^i {{n-1}\choose{i}} \left( \frac{n-1-i}{n} \right)^{k-1} k (k+1)$$?
I have tried changing the order of summation and the getting closed form for the geometric series in $k$ but the terms are large (even for $n = 100$) and I run into precision problems.

Recall that $${n \brace k}=\frac{1}{k!}\sum _{i=0}^k(-1)^i\binom{k}{i}(k-i)^n,$$
so your expression is equal to
$$\sum _{k\geq n}\frac{(n-1)!k(k+1)}{n^{k-1}}{k-1\brace n-1}=2(n-1)!\sum _{k\geq n}\frac{1}{n^{k-1}}\binom{k+1}{2}{k-1\brace n-1},$$
if $n$ is large, then $k$ is large, so ${k-1\brace n-1}\sim \frac{(n-1)^{k-1}}{(n-1)!}$ and so
$$2(n-1)!\sum _{k\geq n}\frac{(n-1)^{k-1}}{n^{k-1}(n-1)!}\binom{k+1}{2}=2\sum _{k\geq n}\left (\frac{n-1}{n}\right )^{k-1}\binom{k+1}{2}$$
now take $x=(n-1)/n,$ we complete the sum to $\infty$ and take out the finite sum getting
$$2\left (\sum _{k=0}^{\infty}\binom{k+1}{2}x^{k-1}-\sum _{k=0}^{n-1}\binom{k+1}{2}x^{k-1}\right )=2\left (\frac{1}{(1-x)^3}-\sum _{k=0}^{n-1}\binom{k+1}{2}x^{k-1}\right ),$$
recall too that taking the second derivative of  $\sum _{k=1}^{n}x^{k}$ gives you
$\sum _{k=1}^{n-1}(k+1)k\cdot x^{k-1}$ but that sum is $\frac{x^{n+1}-1}{x-1}$ and so we get
$$\frac{2}{(1-x)^3}-\frac{(n + 1)n\cdot x^{n-1}}{x-1} + \frac{2 (n+1) x^n}{(x-1)^2} - \frac{2 (-1 + x^{1 + n})}{(x-1)^3},$$
taking $x=(n-1)/n$ back we get
$$2n^3+(n + 1)n^2\cdot \left (1-\frac{1}{n}\right )^{n-1} + 2 n^2(n+1) \left(1-\frac{1}{n}\right )^n - 2 n^3(-1 + \left (1-\frac{1}{n}\right )^{n+1}).$$
For $n$ as big, if I am not mistaken, your expression should be close to
$$(4-\frac{2}{e})n^3+\frac{3}{e}(n+1)n^2$$
