Coupon Collector's Sibling It is given that the coupon collector finishes his collection on average at time $\mathbb{E}(C) = rH_r$, where $H_r = \sum_{i=1}^n \frac{1}{i}$. Let's say that the coupon collector gives all of his duplicate coupons to his little sibling. When the coupon collector finishes, how far is the little sibling from finishing their collection?
It is given that the little sibling is $H_r$ away from finishing their collection in Flajolet. In another paper, I read that the expected number of duplicates for each coupon is given by $P_2^r = \sum_{i=1}^r \frac{i-1}{i}$, which would then give us our answer since $r - \sum_{i=1}^r \frac{i-1}{i}= H_r$, but I don't know how to go about proving that. If anyone could explain where $\sum_{i=1}^r \frac{i-1}{i}$ comes from, that would help immensely in solving this problem.
Edit: Since I believe my friends and I have a constructive proof, if someone could provide a proof using EGF's that would also be cool.
 A: This  happens to  be  very  similar to  a  problem  that has  recently
appeared        here,        namely         at        this        MSE
link.  We  ask  the
reader  to  consult  this  link for  an  introduction  and  additional
background material. 
We start with the species of  ordered set partitions with sets of more
than two elements marked. This is
$$\mathfrak{S}(\mathcal{U}\mathcal{Z}
+\mathcal{U}\mathcal{V}\mathfrak{P}_{\ge 2}(\mathcal{Z})).$$
We thus obtain the generating function
$$G(z, u, v) = \frac{1}{1-u(v\exp(z)-vz+z-1)}.$$
We then get for the probability that
$$P[T=m] = \frac{1}{n^m} {n\choose n-1}
(m-1)! [z^{m-1}] [u^{n-1}] G(z, u, v).$$
What happens here is very simple.  We choose the $n-1$ coupons that go
into the  prefix consisting of  $m-1$ draws.  Then we  partition those
draws into sets, one for each  type of coupon, containing the position
where it appeared.  We mark sets of more than  two elements. Doing the
extraction in $u$ we find
$$P[T=m] = \frac{1}{n^m} {n\choose n-1}
(m-1)! [z^{m-1}] (v\exp(z)-vz+z-1)^{n-1}.$$
Now  to  do  the  usual  sanity  check  that  we  have  a  probability
distribution we remove the marking in $v$ and obtain
$$\sum_{m\ge 1} P[T=m]
= \sum_{m\ge 1} \frac{n!}{n^m}
(m-1)! [z^{m-1}] \frac{(\exp(z)-1)^{n-1}}{(n-1)!}.$$
This  was evaluated  at  the  cited link  and  the  sanity check  goes
through,  more or  less by  inspection  in fact.  Continuing with  the
expectation of coupons that were  drawn mor than once we differentiate
with respect to $v$ and set $v=1$, getting
$$\frac{n! \times (m-1)!}{n^m}
[z^{m-1}] (n-1) \left.\frac{(v\exp(z)-vz+z-1)^{n-2}}{(n-1)!}
\times (\exp(z)-z)\right|_{v=1}
\\ = \frac{n! \times (m-1)!}{n^m}
[z^{m-1}] \frac{(\exp(z)-1)^{n-2}}{(n-2)!} \times (\exp(z)-z).$$
We write this in three pieces, namely
$$\frac{n! \times (m-1)!}{n^m}
[z^{m-1}] \frac{(\exp(z)-1)^{n-1}}{(n-2)!}
\\ - \frac{n! \times (m-1)!}{n^m}
[z^{m-2}] \frac{(\exp(z)-1)^{n-2}}{(n-2)!}
\\ + \frac{n! \times (m-1)!}{n^m}
[z^{m-1}] \frac{(\exp(z)-1)^{n-2}}{(n-2)!}.$$
Consulting the results from the main link we find for the first two pieces
$$n-1 - (H_n - 1) = n - H_n.$$
We then get  for the third piece (recognizing the  Stirling number EGF
and observing that the EGF morphs into an OGF)
$$\frac{n!}{n} \sum_{m\ge 1} \frac{1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-2} \frac{z}{1-qz}
= \frac{n!}{n} \prod_{q=1}^{n-2} \frac{1/n}{1-q/n}
\\ = \frac{n!}{n} \prod_{q=1}^{n-2} \frac{1}{n-q}
= \frac{n!}{n} \frac{1}{(n-1)!} = 1.$$
We  thus have  for  the  answer that  the  sibling collects  $n+1-H_n$
coupons and  hence is  missing $H_n-1$ coupons  probabilistically from
among  the   coupons  collected   in  the  prefix.  Furthermore  and
deterministically, the  sibling never  sees the last  coupon collected
because it is always a singleton. Hence the sibling is missing
$$\bbox[5px,border:2px solid #00A000]{H_n}$$
coupons. We may add the halting  singleton because it does not involve
any additional probability  and is determined by the  set partition of
the prefix.
What have we learned? On seeing this result it immediately becomes
evident  that these  two  parameters (singletons  and duplicates)  are
prefectly additive on  the level of generating functions  and we could
have concluded  by inspection, citing  the result for  singletons from
the link without any extra calculation.
A: I believe some friends of mine and I found an answer. Let $S_i$ be the random variable where $S_i = 1$ if the sibling of the coupon collector does not get the coupon, and let $S_i = 0$ if the sibling of the coupon collector does get the coupon. Line up the coupons in a row beginning from 1 to $r$ in the order in which the coupon collector gets the coupons. Then we have
$$ 1, \ldots, i, \ldots, r.$$
We want to then find $P(S_i = 1)$. Notice that we have $r-i+1$ different places in which the $i$ coupon could appear. Also note that if the $i$ coupon appears after the $r$ coupon, then it does not count, since the coupon collector terminates the sequence after finishing his collection. So, in order to find $P(S_i = 1)$, we simply consider the case in which the $i$ coupon appears after the $r$, which is simply $\frac{1}{r-i+1}$. Note that $S = S_1 + \cdots + S_r$, and by the linearity of expectance $\mathbb{E}(S) = \mathbb{E}(S_1) + \cdots + \mathbb{E}(S_r)$. Hence, to find the expected value, we sum over all different possible $i$'s to get 
$$\mathbb{E}(S) = \sum_{i=1}^r E(S_i) = \sum_{i=1}^r \frac{1}{r-i+1}. $$
Let $r-i+1 = j$. Then we have $$ \mathbb{E}(S) = \sum_{j=1}^r \frac{1}{j},$$
or in other words the expected amount of coupons she is missing is $H_r = \sum_{i=1}^r \frac{1}{i}$.
If any of this looks wrong let me know and I will think about it longer.
