TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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For a nonincreasing nonnegative sequence, $\sum a_k$ converges iff $\sum 2^k a_{2^k}$ converges

Statement

Let $(a_k)_{k \ge 1}$ be a family from $1$ (Series, partial sums, convergence and the sum, divergence, and the tail series) with

$a_k \ge 0 \ \ (k \ge 1) \qquad \text{and} \qquad a_j \ge a_k \ \ \text{whenever } 1 \le j \le k .$

Then

$\sum_{k \ge 1} a_k \ \text{converges} \qquad \Longleftrightarrow \qquad \sum_{j \ge 0} 2^{j} a_{2^{j}} \ \text{converges} .$

Every term of the condensed series is defined, because $2^{j} \ge 1$ for every $j \in \mathbb{N}$ (Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ ), and the condensed series starts at $j = 0$ , its first term being $2^0 a_{2^0} = a_1$ .

The monotonicity hypothesis is equivalent to the consecutive form $a_{k+1} \le a_k$ for every $k \ge 1$ , since it says that the sequence $b_i := a_{i+1}$ , $i \in \mathbb{N}$ , is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences). It cannot be dropped: the companion page exhibits a nonnegative non-monotone family for which the two series behave differently.

Facts & Assumptions

Given: A family $(a_k)_{k \ge 1}$ of reals with $a_k \ge 0$ for $k \ge 1$ and $a_j \ge a_k$ whenever $1 \le j \le k$ ; the partial sums $A_n = \sum_{k=1}^{n} a_k$ of $\sum_{k \ge 1} a_k$ , with $A_0 = 0$ ; and the partial sums $C_n = \sum_{j<n} 2^{j} a_{2^{j}}$ of the condensed series (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Splitting of finite sums, and the meaning of a sum with general bounds: for $m \le n$ , $\sum_{k=1}^{n} a_k = \sum_{k=1}^{m} a_k + \sum_{k=m+1}^{n} a_k$ , and $\sum_{k=p}^{q} a_k = \sum_{t < q+1-p} a_{p+t}$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L2]

Monotonicity and scaling of finite sums: if $x_t \le y_t$ for all $t < d$ then $\sum_{t<d} x_t \le \sum_{t<d} y_t$ ; and a constant sum is $\sum_{t<d} \lambda = d\,\lambda$ (Laws of finite sums and finite products).

[L3]

Powers of $2$ : $2^{j} \ge 1$ for every $j \in \mathbb{N}$ , $2^{j+1} = 2 \cdot 2^{j}$ , and $2^{j} < 2^{j+1}$ (Integer powers $a^m$ , Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ ).

[L4]

The principle of induction (The principle of mathematical induction).

[L5]

For a series of nonnegative terms: its partial sums are nondecreasing, it converges if and only if the range of its partial sums is bounded above, and in the convergent case every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Lower bound, bounded below, bounded set).

Proof

technique · direct

1.1

Every term of the condensed series is nonnegative, since $2^{j} \ge 1 > 0$ and $a_{2^{j}} \ge 0$ ; and every $A_n$ , every $C_n$ is nonnegative, both series having nonnegative terms.

givenL3L5

1.2

For every $n \in \mathbb{N}$ the block from $2^{n}$ to $2^{n+1}-1$ is $\sum_{k=2^{n}}^{2^{n+1}-1} a_k = \sum_{t < 2^{n}} a_{2^{n}+t}$ , since the number of terms is $(2^{n+1}-1)+1-2^{n} = 2^{n}$ .

L1L3algebra

1.3

For every $n \in \mathbb{N}$ the block from $2^{n}+1$ to $2^{n+1}$ is $\sum_{k=2^{n}+1}^{2^{n+1}} a_k = \sum_{t < 2^{n}} a_{2^{n}+1+t}$ , since the number of terms is $2^{n+1}+1-(2^{n}+1) = 2^{n}$ .

L1L3algebra

1.4

Index growth: $n \le 2^{n} - 1$ for every $n \in \mathbb{N}$ , by induction on $n$ . At $n = 0$ this reads $0 \le 0$ ; and if $n \le 2^{n}-1$ , that is $n+1 \le 2^{n}$ , then $n+1 \le 2^{n} \le 2 \cdot 2^{n} - 1 = 2^{n+1}-1$ , the second inequality being $2^{n} \ge 1$ .

L3L4algebra

2.1

In the first block every index satisfies $2^{n} \le 2^{n}+t$ , so $a_{2^{n}+t} \le a_{2^{n}}$ , and therefore $\sum_{k=2^{n}}^{2^{n+1}-1} a_k \le \sum_{t<2^{n}} a_{2^{n}} = 2^{n} a_{2^{n}}$ .

step 1.2givenL2

2.2

In the second block every index satisfies $2^{n}+1+t \le 2^{n+1}$ for $t < 2^{n}$ , so $a_{2^{n}+1+t} \ge a_{2^{n+1}}$ , and therefore $\sum_{k=2^{n}+1}^{2^{n+1}} a_k \ge \sum_{t<2^{n}} a_{2^{n+1}} = 2^{n} a_{2^{n+1}}$ .

step 1.3givenL2L3

2.3

Suppose the condensed series converges, with sum $C$ ; then $C_n \le C$ for every $n$ .

givenstep 1.1L5

2.4

Suppose conversely that $\sum_{k \ge 1} a_k$ converges, with sum $A$ ; then $A_n \le A$ for every $n$ .

givenstep 1.1L5

3.1

Upper estimate: $A_{2^{n}-1} \le C_n$ for every $n \in \mathbb{N}$ , by induction on $n$ . At $n = 0$ both sides are $0$ , since $2^0 - 1 = 0$ and $C_0$ is the empty sum; and if $A_{2^{n}-1} \le C_n$ , then splitting at $2^{n}-1$ gives $A_{2^{n+1}-1} = A_{2^{n}-1} + \sum_{k=2^{n}}^{2^{n+1}-1} a_k \le C_n + 2^{n} a_{2^{n}} = C_{n+1}$ .

step 2.1L1L3L4

3.2

Lower estimate: $2 A_{2^{n}} \ge \sum_{j=1}^{n} 2^{j} a_{2^{j}}$ for every $n \in \mathbb{N}$ , by induction on $n$ . At $n = 0$ the right-hand side is the empty sum $0$ and the left-hand side is $2a_1 \ge 0$ ; and if the inequality holds at $n$ , then splitting at $2^{n}$ gives $A_{2^{n+1}} = A_{2^{n}} + \sum_{k=2^{n}+1}^{2^{n+1}} a_k \ge A_{2^{n}} + 2^{n} a_{2^{n+1}}$ , whence $2A_{2^{n+1}} \ge 2A_{2^{n}} + 2^{n+1} a_{2^{n+1}} \ge \sum_{j=1}^{n+1} 2^{j} a_{2^{j}}$ .

step 2.2step 1.1L1L3L4

4.1

For every $N \in \mathbb{N}$ we have $N \le 2^{N}-1$ , so $A_N \le A_{2^{N}-1} \le C_N \le C$ , the first inequality because the partial sums $A_n$ are nondecreasing.

step 1.4step 3.1step 2.3step 1.1L5

4.2

For every $n \in \mathbb{N}$ , $\sum_{j=1}^{n} 2^{j} a_{2^{j}} \le 2 A_{2^{n}} \le 2A$ , and splitting the condensed partial sum at $j = 1$ gives $C_{n+1} = a_1 + \sum_{j=1}^{n} 2^{j} a_{2^{j}} \le a_1 + 2A$ .

step 3.2step 2.4L1L3

5.1

So the partial sums of $\sum_{k \ge 1} a_k$ are bounded above by $C$ , and that series converges.

step 4.1L5

5.2

Also $C_0 = 0 \le a_1 + 2A$ , so every condensed partial sum is at most $a_1 + 2A$ , and the condensed series converges.

step 4.2step 1.1L5

6.1

The two implications just established combine, so the two series converge or diverge together.

step 5.1step 5.2∎

Remarks

What monotonicity buys, in one sentence. It lets a block of $2^{n}$ consecutive terms be squeezed between $2^{n}$ copies of its last term and $2^{n}$ copies of its first, which is exactly the pair of estimates in steps 2.1 and 2.2. Without it a block carries no information about any single term in it, and the two series decouple entirely.
The factor $2$ in the lower estimate is not an artefact. The blocks used for the two estimates are different: the upper estimate groups $2^{n}, \dots, 2^{n+1}-1$ and the lower estimate groups $2^{n}+1, \dots, 2^{n+1}$ , and the second grouping produces $2^{n} a_{2^{n+1}}$ , which is half of the condensed term $2^{n+1} a_{2^{n+1}}$ . Since only boundedness of the partial sums is at stake, a constant factor is harmless.
Base $2$ is a choice, not a necessity. The same argument with blocks of length $c^{n}$ gives the analogous test for any integer $c \ge 2$ . Base $2$ is taken here because it is the one every later application uses, and because the arithmetic of $2^{n+1} = 2 \cdot 2^{n}$ keeps the induction free of extra bookkeeping.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 84 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Cauchy condensation test (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (3.27) (standard reference, not scraped)
Stephen Semmes, Elements of Analysis (standard reference, not scraped)

For a nonincreasing nonnegative sequence, ∑ak\sum a_k∑ak​ converges iff ∑2ka2k\sum 2^k a_{2^k}∑2ka2k​ converges