Let $U_1,U_2,...,U_n$ be independent random variables that are uniformly distributed in $[0,1]$. Let $N=min$ {$k |U_1+...+U_k>1$}. $(a)$ Let $a∈[0,1]$. Show that $P(U_1+···+U_n≤a)=\frac{a^n}{n!}$. $(b)$ Find $E[N]$ and $Var[N]$.
For $(a)$ I thought about finding the density with convolution; for the general case with $U_1,...,U_n$ I thought about induction. Another idea was to use the moment generating function, but that wasn't very clear to me. How can I approach this exercise? For $(b)$ I don't have any clue. Any help would be much appreciated, I'm not looking for the entire solution for you to give.
Hints (assuming you have an infnite sequence of $U_n$):
a) Using Induction is a good idea. For the inductive step, first calculate the density of $P(U_1 + \dots + U_n)$, and set $X := U_1 + \dots + U_n$. Calculate the density of $X + U_{n+1}$ by taking the convolution of the densities of these two random variables.
b) You have $$\mathbb{E}[N] = \sum_{n=0}^{\infty} P(N > n).$$ and we have $P(N>n) = P(U_1 + \dots + U_n \leq 1)$. Now use part (a).
c) Use $\mathbb{E}[(N-\mu)^2] = \mathbb{E}[N^2] - \mu^2$. Use $P(N=n) = P(N > n-1) - P(N>n+1)$ to calculate $\mathbb{E}[N^2]$.