 Solution

Since the positive series converges, $a_n\to 0$. We construct the subsequence recursively. Choose $n(1)$ arbitrarily. Once $n(1)<\cdots <n(k-1)$ have been chosen, write the reduced partial sums asChoose $n(k)>n(k-1)$ so far out thatThis is possible because $a_n\to 0$.

LetFor every fixed $j$, each later choice includes $q_j^{-j}$ in its minimum, so

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 givecontrary 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 $c>0$ withFor $j>d$ our construction instead giveswhich contradicts the lower bound once $2^{-j}<c$. Hence $\alpha$ is transcendental. This proves the transcendental subseries of a positive rational convergent series construction.