Let $E_n=$ expected number of uniformly random real numbers from $0$ to $1$ needed for their sum to exceed $n$.
Using the probability density function of the Irwin-Hall distribution, I worked out that:
$$E_n=\sum_{i=n}^\infty{\left(\sum_{j=0}^{n}{\frac{(-1)^ji(i-n)(n-j)^{i-1}}{j!(i-j)!}}\right)}$$
which has a closed form in terms of $e$:
$$E_n=\sum_{k=0}^{n}{\frac{(k-n)^k}{k!}e^{n-k}}$$
So we have:
$E_1=e=2.71828...$
$E_2=e^2-e=4.67077...$
$E_3=e^3-2e^2+\frac{1}{2}e=6.66657...$
$E_4=e^4-3e^3+2e^2-\frac{1}{6}e=8.666604...$
$E_5=e^5-4e^4+\frac{9}{2}e^3-\frac{4}{3}e^2+\frac{1}{24}e=10.6666621...$
$E_6=e^6-5e^5+8e^4-\frac{9}{2}e^3+\frac{2}{3}e^2-\frac{1}{120}e=12.6666671...$
etc.
It is evident that $E_n\to2n+\frac{2}{3}$ as $n\to\infty$. My question is, how can we prove that:
$$\lim_{n\to\infty}(E_n-2n)=\frac{2}{3}$$
(Context: I was trying to generalize this question.)
EDIT: I just found that this is essentially a duplicate of this answered question. The link appeared in the "Related" question list on the right side.
