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Source. Example 3.1 and its proof, preprint p. 4. Read on the rendered page.

Statement

Example 3.1 (p. 4): "For any integer $k>0$ the sum

$$\sum_{n=1}^{\infty}\frac{p_n^k}{2^{p_n}}$$

is irrational."

Proof sketch (p. 4)

The example is an application of Theorem 3.1 with $b_n=p_n^k$ and $a_n=2^{p_n-p_{n-1}}$, where $p_0=0$. The products $a_1\cdots a_n$ are then $2^{p_n}$, so $S$ is the series above, and every $a_n$ is at least $2$ because consecutive primes differ. The growth hypothesis on $b_n$ holds for large $n$ because $p_{n+1}/p_n\to1$ (the print states it as $p_{n+1}^k-p_n^k<\epsilon p_n^k$). For $\liminf b_n/a_n=0$ the paper cites the theorem of Westzynthius [12] that $\limsup_{n\to\infty}(p_{n+1}-p_n)/\log p_n=\infty$. It gives infinitely many $n$ whose preceding gap $p_n-p_{n-1}$ exceeds $2k\log n$. At those $n$ the denominator $a_n$ exceeds a power of $n$ with exponent above $1.2k$, while $p_n^k$ is at most $(n(\log n)^2)^k$, so $b_n/a_n\to0$ along them.

The paper closes the example with: "The case $k=1$ is claimed by Erdős and Graham [4], page 62." The proof is complete given Theorem 3.1 and Westzynthius's result, cited to [12].

Relation to problem 251

Problem 251 concerns $\sum p_n/2^n$, with the prime at position $n$; here the prime sits at position $p_n$, so the series is much sparser and the argument is elementary. The example is adjacent to the problem and is not progress on it. The attribution of the case $k=1$ to Erdős and Graham 1980, p. 62, is the paper's. Printed p. 62 of that monograph (card erdos_1980_old_new_problems_results_combinatorial_number_theory) names no prime series of this shape, but says that $\sum_na_n/2^{a_n}$ "is known to be irrational under the stronger hypothesis that $a_n>cn\sqrt{\log n\log\log n}$", without a proof or a reference; $a_n=p_n$ satisfies that hypothesis for large $n$, since $p_n\sim n\log n$, so the claim covers the case $k=1$.

Bears on. #251 (adjacent series; a mention that explains the difference).