 Solution

Choose a subsequence $a_{n_j}$ so sparse thatfor every $j$; this is possible because the tails of the convergent positive series tend to zero. Distinct binary sequences $({\epsilon }_j)$ then have distinct sums ${\sum }_j{\epsilon }_j{a}_{n_j}$: at their first disagreement, the corresponding term exceeds the entire remaining tail. Thus these subsums form an uncountable set. The algebraic numbers are countable, so at least one subsum is transcendental. Set ${\theta }_{n_j}={\epsilon }_j$ and all other ${\theta }_n=0$.