Let $N(x)$ be the number of uniform random variables (distributed in $[0,1]$) that one needs to add for the sum to cross $x$ ($x > 0$). The expected value of $N(x)$ can be calculated and it is a very cool result that $E(N(1)) = e$. The expression for general $x$ is 


$E(N(x)) = \sum\_{k=0}^{[x]} (-1)^k \frac{(x-k)^k}{k!} e^{x-k}$


where $[x]$ is the largest integer less than $x$. When $x$ is large, this function $E(N(x))$ grows linearly (as one would expect). Computer simulations suggest that $E(N(x)) \approx 2x + 2/3$. My question is as follows: is it possible to guess this form for $E(N(x))$ without actually computing it? The $2x$ part is intuitive but is there a good intuition for why there is an additive constant and why that value is $2/3$?


My motivation is as follows: When I computed $E(N(x))$ using the formula above for large values of $x$, I came across this asymptotic and found it surprising that a polynomial equation in powers of $e$ gives values that are very close to a rational number. I am very interested in knowing the reason (if any) behind it. Thanks in advance for any help.



Number of uniform rvs needed to cross a threshold
Another way to put it: the expected value of the sum right after it crosses x is x+1/3. If you simply conditioned the sum to be in (x,x+1) then the expectation would be about x+1/2, with the sum almost uniformly distributed. But you also condition on the overshoot being less than the last jump. The expectation of the smaller of two uniform [0,1] iids is 1/3.


