There must be some missing constraints. If $\alpha_n$ is allowed to be negative, we get the following counterexample. $\smash{\rlap{\phantom{\Bigg(}}}$
Define $ u_{n+1}=(1-\alpha_n)u_n+\beta_n\tag{1} $ and $ A_n=\prod_{k=1}^{n-1}(1-\alpha_k)\tag{2} $ By induction, it can be verified that $ u_n=A_n\left(u_1+\sum_{k=1}^{n-1}\frac{\beta_k}{A_{k+1}}\right)\tag{3} $ For $j\ge1$, define $ n_j=\left\{\begin{array}{} 2^{j(j-1)/2}&\text{when }j\text{ is odd}\\ 2^{j(j-1)/2+1}&\text{when }j\text{ is even} \end{array}\right.\tag{4} $ and for $n\ge1$, $ \alpha_n=\left\{\begin{array}{} \frac{1}{n+1}&\text{for }n_j\le n< n_{j+1}\text{ when }j\text{ is odd}\\ -\frac1n&\text{for }n_j\le n< n_{j+1}\text{ when }j\text{ is even} \end{array}\right.\tag{5} $ Obviously, $\displaystyle\lim_{n\to\infty}\alpha_n=0$.
Using telescoping products, it is not difficult to show that $ \frac{A_{n_{j+1}}}{A_{n_j}}=\left\{\begin{array}{} \frac{n_j}{n_{j+1}}=2^{-j-1}&\text{when }j\text{ is odd}\\ \frac{n_{j+1}}{n_j}=2^{j-1}&\text{when }j\text{ is even} \end{array}\right.\tag{6} $ Equation $(6)$ yields $ A_{n_j}=\left\{\begin{array}{} 2^{-(j-1)/2}&\text{when }j\text{ is odd}\\ 2^{-(3j-2)/2}&\text{when }j\text{ is even} \end{array}\right.\tag{7} $ Furthermore, using the standard formula for the partial harmonic series, when $j$ is odd, $ \begin{align} \sum_{n=n_j}^{n_{j+1}-1}\alpha_n &=\log\left(\frac{n_{j+1}}{n_j}\right)+O\left(\frac{1}{n_j}\right)\\ &=(j+1)\log(2)+O\left(2^{-j(j-1)/2}\right)\tag{8} \end{align} $ and when $j$ is even, $ \begin{align} \sum_{n=n_j}^{n_{j+1}-1}\alpha_n &=-\log\left(\frac{n_{j+1}}{n_j}\right)+O\left(\frac{1}{n_j}\right)\\ &=-(j-1)\log(2)+O\left(2^{-j(j-1)/2}\right)\tag{9} \end{align} $ Combining $(8)$ and $(9)$ yields $ \sum_{n=1}^{n_j-1}\alpha_n=\left\{\begin{array}{} \frac{j-1}{2}\log(2)+O(1)&\text{when }j\text{ is odd}\\ \frac{3j-2}{2}\log(2)+O(1)&\text{when }j\text{ is even} \end{array}\right.\tag{10} $ Equation $(10)$ says that $\displaystyle\sum_{n=1}^\infty\alpha_n=\infty$.
Define $ \beta_n=\left\{\begin{array}{} 2^{-j}&\text{when }n=n_j-1\text{ for }j\text{ even}\\ 0&\text{otherwise} \end{array}\right.\tag{11} $ Summing the geometric series yields $\displaystyle\sum_{n=1}^\infty\beta_n=\frac13$.
Using $(3)$, we get $ \begin{align} u_{n_{j+1}} &=A_{n_{j+1}}\left(u_1+\sum_{k=1}^{n_{j+1}-1}\frac{\beta_k}{A_{k+1}}\right)\\ &\ge\frac{A_{n_{j+1}}}{A_{n_j}}\beta_{n_j-1}\\ &=2^{j-1}\cdot2^{-j}\\ &=\frac12\tag{12} \end{align} $ when $j$ is even. $(12)$ says that $\displaystyle\lim_{n\to\infty}u_n\not=0$.