Is there any way to simplify the following expression?
$ \sum_{d = 1}^k \left(\sum_{i=1}^d \frac{1}{i}\right) \frac{{n-t \choose d}{t \choose k-d}}{{n \choose k}} $
This formula comes from the expected number of record lows over the first $k$ elements in a permutation of $[1,n]$, given some minimum threshold $t$ below which the elements don't count as a record low.
Let $L$ be the number of record lows, and let $d$ be the number of elements above the threshold. $ E[L] = E[E[L|d]] = \sum_{d=1}^k E[L|d]\cdot P(d) $
where $P(d)$ is a hypergeometric distribution, with n-t success states, population size $n$, and $k$ draws.
This comes up for example in this question: Expected number of cards in the stack?