The closed form of the expected number of samples for $t+1>\sum_r X_r \geqslant t, X_r \sim \text{U[0,1]}$ is given by:
$$m(t) = \sum_{k=0}^{\lfloor t \rfloor} \frac{(k-t)^k}{k!}e^{t-k}$$
From this we can deduce the expected amount by which this sum exceeds $t$, namely:
$$\varepsilon(t) = \frac{m(t)}{2} - t$$
From knowing that $m(t) \to \dfrac{t}{\mathbb{E}[X]}+\dfrac{\mathbb{E}[X^2]}{2\mathbb{E}[X]^2}$, we can easily see that $\varepsilon(t) \to \dfrac{1}{3}$.
Is there a simple ("low tech") way of proving that $\varepsilon(t) \to \dfrac{1}{3}$ without first passing through proving $m(t) \to 2t+\dfrac{2}{3}$ ?