Adding until past one

Idea. Turn the expectation into a sum of tail probabilities. Each tail probability is the volume of a simplex, which is $\frac{1}{n!}$, and the sum of those is $e$.

Step 1: tail sum. For a variable with values in $\{1,2,\dots \}$, $$\mathbb{E}[N]=\sum _{n=0}^{\infty }\mathbb{P}(N>n),$$ and $\{N>n\}=\{U_1+\cdots +U_n\le 1\}$. For $n=0$ the empty sum is $0\le 1$, so that term is $1$.

Step 2: volume of the simplex. $\mathbb{P}(U_1+\cdots +U_n\le 1)$ is the volume of $S_n=\{x\in [0,1{]}^n:x_1+\cdots +x_n\le 1\}$. The map $(x_1,\dots ,x_n)\mapsto (x_1,x_1+x_2,\dots ,x_1+\cdots +x_n)$ has determinant $1$ and sends $S_n$ onto the set of increasing sequences $0\le y_1\le \cdots \le y_n\le 1$. That set is one of the $n!$ equally sized pieces of the cube obtained by ordering the coordinates, so its volume is $\frac{1}{n!}$.

Step 3: sum. $$\mathbb{E}[N]=\sum _{n=0}^{\infty }\frac{1}{n!}=e\approx \mathrm{2.718.}$$

Sanity check. $N$ is at least $2$, and $\mathbb{P}(N=2)=1-\frac{1}{2}=\frac{1}{2}$, so an average a bit above $2.5$ is plausible.