expected value of sum of random variables 
Suppose $U_i\sim U(0,1)$ for $1\leq i\leq n$ independently. Define the random variable $T = \min \{t : \sum_{i=1}^t U_i \geq c\}$, where $c>0$. What is the expected value of $T$? What does this value approach as $k\to \infty$?

I think $P(T > t) = P(\sum_{i=1}^t U_i < c)$ and $\sum_{t=0}^\infty P(T > t) = \sum_{t=0}^\infty \sum_{x=t}^\infty P(T =x) = \sum_{x=0}^\infty \sum_{t=0}^x P(T = x) = \sum_{x=0}^\infty xP(T = x) = E(T).$
Also, by the central limit theorem, $\sum_{i=1}^t U_i$ approaches the $N(0,t)$ normal distribution as the $U_i$ are independent. So would it be correct to say that $E(T) = \sum_{t=0}^\infty P(\sum_{i=1}^t U_i<c)$ approaches $\sum_{t=0}^\infty P(0 \leq Z < \frac{c}{\sqrt{t}})$ for $c$ large? If so, how could one simplify this and if not, is there an easier method?
 A: If we denote $S_t=\sum_i^tU_i$, we can rewrite your summation as:
$$
\mathbb{E}T=f(c)=\sum_{t=0}^\infty P(S_t<c).
$$
Let's find out $f(c)$ when $0<c<1$. Condition $\sum U_i<c$ and $U_i>0$ gives us $t$-dimensional simplex in the metric space $(U_1,\ldots U_t)$. Since $c<1$, $U_i<1$ are true for all points of the simplex. And thus $P(S_t<c)$ is equal the ratio of volumes of simplex and cube $[0,1]^t$, which is $c^t/t!$ Thus,
$$
P(S_t<c) = \sum\frac{c^t}{t!}=e^c,\qquad\text{when}\quad 0<c<1.
$$
If $c>1$, then with one jump we surely don't reach $c$. So we can perform one jump by $x$ and then we will need to find the expected number of jumps to reach $c-x$:
$$
f(c) = 1 + \mathbb{E}_x f(c-x) = 1 + \int_0^1f(c-x)dx=1+\int_{c-1}^c f(z) dz
$$
If we denote as $f_{n}(c)=f(c)$ when $c\in[n, n+1]$, $n\in\mathbb{N}$, then after differentiation by $c$:
$$
f'_n(c)=f_n(c)-f_{n-1}(c-1),
$$
with continuity condition that $f_n(n) = f_{n-1}(n)$.
Since $f_0(c)=e^c$, we can make an educated guess to have $f_n(c)=P_0(c)e^c+P_1(c)e^{c-1}+\ldots P_n(c)e^{c-n}$, where $P_n(c)$ is a polynomial of power $n$. Indeed:
$$
f'_n(c) = \left(P_0(c)e^c+P_1(c)e^{c-1}+\ldots P_n(c)e^{c-n}\right) + \left(P'_0(c)e^c+P'_1(c)e^{c-1}+\ldots P'_n(c)e^{c-n}\right) =
\left(P_0(c)e^c+P_1(c)e^{c-1}+\ldots P_n(c)e^{c-n}\right) - 
\left(P_0(c-1)e^{c-1}+P_1(c-1)e^{c-2}+\ldots P_{n-1}(c-1)e^{c-n}\right),\\
P'_1(c)e^{-1}+\ldots P'_n(c)e^{-n} = -\left(P_0(c-1)e^{-1}+\ldots P_{n-1}(c-1)e^{-n}\right).
$$
If we let $P_n(c)=\frac1{n!}(n-c)^n$,
then
$$
P'n(c) = \frac n{n!}(n-c)^{n-1} = -\frac 1{(n-1)!}\Big((n-1)-(c-1)\Big)^{n-1} = -P_{n-1}(c-1).
$$
Thus, final formula is:
$$
f(c) = \sum_{n=0}^{\lceil c\rceil}\frac{(n-c)^n}{n!}e^{c-n}.
$$
When $c\to\infty$, we can notice that for $t<c$, $P(S_t<c)=1$ and for $t>c$ $S_t\sim N(t/2, t/12)$, thus $P(S_t<c)=1-\Phi(\frac6t(2c-t))$. So the relative number of $t$, where $P(S_t<c)$ is significantly different from 1 or 0 goes down as $1/\sqrt{t}$, and finally $f(c)\approx 2c$
