Take $g_i$ to be a geometric random variable with parameter $1/2$, such that $$P(g_i = k) = \frac{1}{2^{k+1}}$$ for any integer $i$.
I'm surprised at how much more difficult it is to evaluate this sum (of independent variables)
$$\sigma \sim \sum_{i=1}^{N}ig_i = g_1 + 2g_2 + \dots + Ng_N$$
than this simpler, similar sum (also of independent variables)
$$\nu \sim \sum_{i=1}^{N}g_i = g_1 + g_2 + \dots + g_N$$
I understand that the distribution of $\sigma$ is related to the number of lattice paths from $(0,0)$ to $(N,k)$ which take only north or east steps and enclose an area $n$, summed over all $k$, i.e.
$$P(\sigma = n) = \sum_{k=1}^{\infty}\frac{1}{2^{N+k}}p(n,\{k\},N)$$
where $p(n,\{k\},N)$ is the number of integer partitions of $n$ into exactly $k$ parts none of which are larger than $N$. This is becuase $g_1 + 2g_2 + \dots + Ng_N$ is actually a Ferrer's diagram (whose boundary is a lattice path). Alternatively, write the generating function of $\sigma$ $$\mathbb{E}(z^{\sigma})=\prod_{i=1}^{N}\frac{1}{2-z^i}$$ and ask for the coefficient of $z^n$, which is $P(\sigma = n)$. I cannot find a closed form for this even in a limit. Can anyone advise?
Also, if $\nu$ has the negative binomial distribution, perhaps so does $\sigma$? If not, is there some limit where this is the case?
