= Solution
Since the positive series converges, $a_n\to0$. We construct the subsequence recursively. Choose $n(1)$ arbitrarily. Once $n(1)<\cdots<n(k-1)$ have been chosen, write the reduced partial sums as
$$
s_j=\sum_{r=1}^j a_{n(r)}=\frac{p_j}{q_j}
\qquad(1\leq j<k).
$$
Choose $n(k)>n(k-1)$ so far out that
$$
0<a_{n(k)}<2^{-k}min_{1\leq j<k}q_j^{-j}.
$$
This is possible because $a_n\to0$.
Let
$$
\alpha=\sum_{k=1}^{\infty}a_{n(k)}.
$$
For every fixed $j$, each later choice includes $q_j^{-j}$ in its minimum, so
$$
0<\alpha-s_j
<q_j^{-j}\sum_{k=j+1}^{\infty}2^{-k}
=2^{-j}q_j^{-j}.
$$
First, $\alpha$ is irrational. If $\alpha=u/v$ were reduced and rational, the strictly increasing rational sequence $s_j\to\alpha$ would have unbounded denominators $q_j$; only finitely many reduced fractions in a bounded interval have bounded denominator. Choose $j>2$ with $q_j>v$. Part (b) would give
$$
|\alpha-s_j|>q_j^{-2},
$$
contrary to $|\alpha-s_j|<2^{-j}q_j^{-j}<q_j^{-2}$.
If $\alpha$ were algebraic of degree $d$, part (a) would give a constant $c>0$ with
$$
|\alpha-s_j|>c q_j^{-d}.
$$
For $j>d$ our construction instead gives
$$
|\alpha-s_j|<2^{-j}q_j^{-j}
\leq2^{-j}q_j^{-d},
$$
which contradicts the lower bound once $2^{-j}<c$. Hence $\alpha$ is transcendental. This proves the <transcendental subseries of a positive rational convergent series> construction.
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