asymptotics for sums of uniform random variables Let  U(1) , U(2) , ...   be independent uniform random variables in  [0,1] .
For  x > 0  define  N(x) by
N(x)  =  min { n : U(1) + U(2) + ... + U(n) >  x  } .
Finally, let  
E(x)  =  expected value of the random variable  N(x) .
It is known that   E(1) = e = 2.718...  [e.g. Ross , First course 
in probability; 4th ed. , p. 339 ] .
This clearly gives the upper bound E(x) < x*e  for  x >> 0 .
Questions: a) Is there an asymptotic formula for E(x)?
       b) Does the limit of E(x)/x  exist as x tends to infinity?


Conditioning on $U(1)$ and integrating yields 
$$
E(x)=1+\int_{x-1}^xE(y)\mathrm dy,
$$ 
for every $x\geqslant0$ with the initial condition that $E(x)=0$ for every $x\lt0$. This implies the result you mention that $E(x)=\mathrm e^x$ for every $x$ in $[0,1]$, in particular $E(1)=\mathrm e$. Furthermore, considering the Laplace transform 
$$
L(t)=\int_0^\infty E(x)\mathrm e^{-tx}\mathrm dx,
$$
defined for every positive $t$, one can use the integral identity above to compute $L(t)$, yielding 
$$
L(t)=\frac1{\mathrm e^{-t}-1+t}.
$$ 
The asymptotics of $L(t)$ when $t\to0$ yields the asymptotics of $E(x)$ when $x\to\infty$. 
Namely, the expansion of $\mathrm e^{-t}$ up to order $t^3$ yields $$
L(t)=\frac2{t^2}\left(1-\frac13t+O(t^2)\right)^{-1}=\frac2{t^2}+\frac2{3t}+O(1),
$$
which implies that 
$$E(x)=2x+\frac23+O\left(\frac1x\right).$$
Keeping more terms in the expansion of $\mathrm e^{-t}$, one gets refined expansions of $L(t)$, hence the coefficients of the next terms $\frac1x$, $\frac1{x^2}$, and so on, in the expansion of $E(x)$.
