EDIT: for a consolidated answer to the general variance in a coupon collectors problem with unequal probabilities, see here: https://math.stackexchange.com/a/3454032/155881
In example 5.17 of the book on Introduction to probability models by Ross, he solves the coupon collector's problem, where there are $n$ coupons, each with probability $p_j$ of being collected per draw (with $\sum_{j=1}^n p_j=1$). He uses the Poisson process to come up with the following expression for the expected value of $X$, the number of coupons to be collected for completing the collection:
$$E(X) = \int\limits_0^\infty P(X>t)dt = \int\limits_0^\infty \left(1-\prod\limits_{j=1}^n (1-e^{-p_j t})\right)dt$$ Using the fact that $\int_0^\infty e^{-pt}=\frac 1 p$,
$$E(X)=\sum \frac 1 p_j -\sum_{i<j} \frac{1}{p_i+p_j}+\dots +(-1)^{n-1}\frac{1}{p_1+\dots+p_n}$$
Now, I want to use the same approach to calculate the variance. Per comment by @BGM here and also this question, we can use the following expression to get $E(X^2)$:
$$E(X^2) = \int\limits_0^\infty 2tP(X>t)dt = \int\limits_0^\infty 2t\left(1-\prod\limits_{j=1}^n(1-e^{-p_j t})\right)dt$$
Using the fact that $\int\limits_0^\infty te^{-pt}=\frac{1}{p^2}$ and the same algebra as for $E(X)$ we get:
$$\frac{E(X^2)}{2} = \sum \frac {1} {p_j^2} -\sum_{i<j} \frac{1}{(p_i+p_j)^2}+\dots +(-1)^{n-1}\frac{1}{(p_1+\dots+p_n)^2} $$
Now, let's consider the special case where all coupons have an equal probability of being selected. In other words, $p_j=\frac 1 n \; \forall \; j$.
Approach-1 We get:
$$\frac{E(X^2)}{2} = n^2\left(\sum\limits_{k=1}^n (-1)^{k-1}\frac{n\choose k}{k^2}\right)$$
Per my answer to the question here, this summation yields:
$$E(X^2) = 2n^2\left( \sum_{j=1}^n\sum_{k=1}^j\frac{1}{jk}\right)\tag{1}$$
Approach-2 But per this paper, the variance for this special case is:
$$V(X) = n^2\sum_{j=1}^m\frac{1}{j^2}-n\sum_{j=1}^m\frac{1}{j} $$ and this would mean that:
$$E(X^2) = V(X)+E(X)^2 = n^2\sum_{j=1}^m\frac{1}{j^2}-n\sum_{j=1}^m\frac{1}{j}+\left(n\sum_{j=1}^m\frac{1}{j}\right)^2$$
If we visualize a $j-k$ grid, it's easy to see that this is the same as:
$$E(X^2) = 2n^2\left( \sum_{j=1}^n\sum_{k=1}^j\frac{1}{jk}\right)-n\sum_{j=1}^m\frac{1}{j}\tag{2}$$
If we compare equation (1) from approach-1 and equation (2) from approach-2, it's clear that equation (1) has a missing $-n\sum_{j=1}^m\frac{1}{j}$ term. And equation (2) has been verified using other methods. This indicates that there is some small mistake with approach-1 that is making us miss this term. I haven't been able to spot what this issue is. Hoping someone else might.
