I am trying to evaluate the following finite sum:
$$
\sum_{k=1}^{n}(-1)^{k}(k-1)!S(n-1, k-1)(\sum_{i=0}^{k-1}H_{i}),
$$
where $S(n, k)$ are the Stirling's numbers of the second kind and $H_{i}$ denotes the $i$ harmonic number. Could you please shed some light?
@Gerry Myerson
I think that the first few terms obey the sequence http://oeis.org/A001787 with alternating signs.
Your OEIS guess is correct. The trick is looking at this formula for Stirling numbers of the second kind:
$$\sum_{k = 0}^n {n \brace k} (x)_k = x^n$$
(I'm using Pochhamer symbols)
and noticing that Pochhamer symbol turns out to be really simple for some special values:
$$ (-1)_n = (-1)^n n! $$ $$ (-2)_n = (-1)^n (n+1)! \ \ \ (*)$$
This gives us a few surprising identities:
$$\sum_{k = 0}^n {n \brace k} (-1)^k k! = (-1)^n$$ $$\sum_{k = 0}^n {n \brace k} (-1)^k (k+1)! = (-2)^n$$
That's nice, but I still haven't solved your question, right? Your OEIS sequence looks like $n 2^{n-1}$ - comparing this and few last identities you see it's a derivative of some kind.
This is how harmonic numbers enter the game. Notice:
$$\frac{d}{dx} (x)_n = (x)_n \sum_{k = 0}^{n-1} \frac1{x-k}$$ (just using the multiplication formula for derivation).
Just differentiate the very first formula I wrote in this post, put $x = -2$ and use identity $(*)$!