A nonnegative non-monotone sequence for which $\sum a_k$ and $\sum 2^k a_{2^k}$ behave differently

Refuted claim: for every family $(a_k)_{k \ge 1}$ with $a_k \ge 0$, $\sum_{k \ge 1} a_k$ converges if and only if $\sum_{j \ge 0} 2^{j} a_{2^{j}}$ converges.

This is For a nonincreasing nonnegative sequence, $\sum a_k$ converges iff $\sum 2^k a_{2^k}$ converges with its monotonicity hypothesis deleted. Let $P := \{2^{j} : j \in \mathbb{N}\}$ be the set of powers of $2$ and define, for naturals $k \ge 1$,

$a_k \;:=\; \begin{cases} 0 & \text{if } k \in P, \\ 1 & \text{if } k \notin P. \end{cases}$

Every term is nonnegative, and the family is not monotone in either direction: $a_1 = 0 < 1 = a_3$ and $a_3 = 1 > 0 = a_4$, since $1 = 2^{0}$ and $4 = 2^{2}$ belong to $P$ while $3$ does not.

The condensed series is $\sum_{j \ge 0} 2^{j} a_{2^{j}} = \sum_{j \ge 0} 0$, which converges with sum $0$. The original series $\sum_{k \ge 1} a_k$ diverges, because $a_k = 1$ at arbitrarily large indices, so its terms do not tend to $0$ (If a series converges then its terms tend to $0$).

• The witness knocks out exactly one estimate. Condensation squeezes the block $a_{2^{n}}, \dots, a_{2^{n+1}-1}$ between $2^{n}$ copies of its last term and $2^{n}$ copies of its first, and both bounds are consequences of monotonicity. Here the first term of each block is $0$ and the rest are $1$, so the upper bound $2^{n} a_{2^{n}} = 0$ is wildly wrong, and it is the upper bound that the convergence direction of the theorem uses.

• The failure is one-directional here, and the other direction can fail too. This witness has a convergent condensed series and a divergent original. The complementary family $a_k := 1/k$ for $k \in P$ and $a_k := 0$ otherwise reverses the roles, its original series being a geometric one and its condensed series having every term equal to $1$; that variant is not verified here, and only the direction exhibited above is claimed.