For $(a_k)$ nonincreasing and nonnegative, $\sum a_k$ converges exactly when $\sum 2^k a_{2^k}$ converges.Suggest a slogan
Let \((a_k)_{k\geq1}\) be a nonincreasing sequence of nonnegative real numbers. Then \[ \sum_{k\geq 1} a_k < \infty \iff \sum_{k\geq 1} 2^k a_{2^k}<\infty . \]
Group the terms in dyadic blocks \(2^k\le j<2^{k+1}\). Since \((a_j)\) is nonincreasing, each block satisfies \[ 2^k a_{2^{k+1}}\le\sum_{j=2^k}^{2^{k+1}-1}a_j\le 2^k a_{2^k}. \] Summing over \(0\le k\le K\) gives \[ \frac12\sum_{k=1}^{K+1}2^{k}a_{2^{k}}\le\sum_{j=1}^{2^{K+1}-1}a_j\le\sum_{k=0}^{K}2^k a_{2^k}. \] Both series have nonnegative terms, so their partial sums are nondecreasing, and these bounds show that one sequence of partial sums is bounded if and only if the other is; the term \(a_1\) omitted from \(\sum_{k\geq1}2^ka_{2^k}\) does not affect convergence.