= Solution
For $\alpha>0$, integration and <independence of random variables> give
$$
\mathbb EU_1^\alpha=\frac1b\int_0^bu^\alpha\,du=\frac{b^\alpha}{1+\alpha},\qquad \mathbb EW_n^\alpha=\left(\frac{b^\alpha}{1+\alpha}\right)^n.
$$
By the <fractional moment criterion for convergence in probability>,
$$
\mathbb P(W_n>\varepsilon)\leq\varepsilon^{-\alpha}\left(\frac{b^\alpha}{1+\alpha}\right)^n.
$$
The parenthesized factor is less than one precisely when $\alpha\log b-\log(1+\alpha)<0$. This expression vanishes at $\alpha=0$ and has <derivative> $\log b-1$ there. If $b<e$, the <derivative> is negative, so some sufficiently small $\alpha>0$ makes the factor strictly less than one. The bound then tends to zero geometrically, proving
$$
\boxed{W_n\longrightarrow0\quad\text{in probability for every }0<b<e.}
$$
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