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

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For a divergent series of positive terms with partial sums $s_k$ , the series $\sum a_k/s_k$ diverges and $\sum a_k/s_k^2$ converges

Statement

Let $(a_k)$ be a sequence of reals with $a_k > 0$ for every $k \in \mathbb{N}$ and suppose $\sum a_k$ diverges (Series, partial sums, convergence and the sum, divergence, and the tail series). Write

$S_n \;:=\; \sum_{k=0}^{n} a_k \qquad (n \in \mathbb{N})$

for the inclusive partial sums, so that $S_n = s_{n+1}$ in the notation $s_n = \sum_{k<n} a_k$ of Series, partial sums, convergence and the sum, divergence, and the tail series. Then $S_n > 0$ for every $n \in \mathbb{N}$ , and:

$\displaystyle\sum_{n} \frac{a_n}{S_n}$ diverges;
$\displaystyle\sum_{n} \frac{a_n}{S_n^{2}}$ converges.

Why the divisor is the inclusive partial sum. The exclusive partial sum of Series, partial sums, convergence and the sum, divergence, and the tail series has $s_0 = 0$ , the empty sum, so $a_0/s_0$ has no value and a series divided by $s_n$ would have to begin at $n = 1$ . The inclusive sum has $S_0 = a_0 > 0$ , so both series above are series of sequences on $\mathbb{N}$ with no shift and no excluded index. The classical statement, which writes $s_n = a_1 + \dots + a_n$ and starts at $n = 1$ , is this one with the indices moved by one.

What the theorem says. No divergent series of positive terms is slowest: dividing its terms by the running total produces a series that still diverges but whose terms are eventually strictly smaller. Dividing by the square of the running total overshoots and produces a convergent series. Claim 1 is what refutes the existence of a universal comparison series (FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test).

Facts & Assumptions

Given: A sequence $(a_k)$ of reals with $a_k > 0$ for every $k$ , with $\sum a_k$ divergent; the exclusive partial sums $s_n = \sum_{k<n} a_k$ and the inclusive partial sums $S_n = \sum_{k=0}^{n} a_k = s_{n+1}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

For a series of nonnegative terms: the partial sums are nondecreasing, the series converges if and only if their range is bounded above, and otherwise they diverge to $+\infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, A nondecreasing sequence that is not bounded above diverges to $+\infty$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Divergence to $+\infty$ and to $-\infty$ ).

[L2]

Splitting of finite sums: for $m \le n$ , $\sum_{k=0}^{n} a_k = \sum_{k=0}^{m} a_k + \sum_{k=m+1}^{n} a_k$ , so $\sum_{k=m+1}^{n} a_k = S_n - S_m$ ; and monotonicity of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L3]

The Cauchy criterion: $\sum d_k$ converges if and only if for every real $\varepsilon > 0$ there is $N$ with $\big|\sum_{k=m+1}^{n} d_k\big| < \varepsilon$ for all $n > m \ge N$ (A series converges iff for every $\varepsilon > 0$ there is $N$ with $|a_{m+1} + \dots + a_n| < \varepsilon$ for all $n > m \ge N$ ).

[L4]

For positive terms, $x_n \to 0$ if and only if $1/x_n \to +\infty$ (For positive terms, null and divergence to $+\infty$ are reciprocal, Limits and Cauchy sequences of reals).

[L5]

$\sum (b_n - b_{n+1})$ converges whenever $(b_n)$ converges ( $\sum (b_k - b_{k+1})$ converges iff $(b_k)$ converges, with sum $b_0 - \lim b_k$ ).

[L6]

Direct comparison (If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$ , and divergence of $\sum a_k$ gives divergence of $\sum b_k$ ), and a series converges if and only if each of its tail series converges (A series converges iff each of its tail series converges, and the sum splits as $s_N$ plus the $N$ -th tail).

[L7]

Reciprocation on the positives: $0 < x \le y$ implies $0 < 1/y \le 1/x$ (Inverses of positives are positive, and reciprocation reverses order); and $x^{2} = x \cdot x$ (Integer powers $a^m$ ).

Proof

technique · direct

1.1

Every $S_n$ is a sum of positive terms, so $S_n > 0$ ; and $S_n \le S_{n+1}$ since $S_{n+1} - S_n = a_{n+1} > 0$ , so $(S_n)$ is nondecreasing and $S_m \le S_n$ whenever $m \le n$ .

givenL2L1

1.2

Since $\sum a_k$ diverges and its terms are nonnegative, the exclusive partial sums are unbounded above and $s_n \to +\infty$ ; hence $S_n = s_{n+1} \to +\infty$ , because for a given real $M$ any index bound $K$ working for $(s_n)$ also works for $(S_n)$ .

givenL1algebra

2.1

For all naturals $n > m$ , using $S_k \le S_n$ for $k \le n$ and $a_k > 0$ : $\displaystyle\sum_{k=m+1}^{n} \frac{a_k}{S_k} \ \ge \ \sum_{k=m+1}^{n} \frac{a_k}{S_n} \ = \ \frac{S_n - S_m}{S_n} \ = \ 1 - \frac{S_m}{S_n}$ .

step 1.1L2L7

2.2

Put $b_n := 1/S_n$ , which is positive; since $1/b_n = S_n \to +\infty$ , the sequence $(b_n)$ converges to $0$ .

step 1.1step 1.2L4L7

2.3

For every $n \in \mathbb{N}$ : $b_n - b_{n+1} = \dfrac{1}{S_n} - \dfrac{1}{S_{n+1}} = \dfrac{S_{n+1}-S_n}{S_n S_{n+1}} = \dfrac{a_{n+1}}{S_n S_{n+1}} \ \ge \ \dfrac{a_{n+1}}{S_{n+1}^{2}} \ > \ 0$ , the inequality because $0 < S_n S_{n+1} \le S_{n+1}^{2}$ .

step 1.1L2L7algebra

3.1

Let $N \in \mathbb{N}$ be arbitrary and put $m := N$ . Since $S_n \to +\infty$ there is $n > m$ with $S_n > 2 S_m$ , and then $S_m/S_n < 1/2$ , so the block of step 2.1 satisfies $\sum_{k=m+1}^{n} a_k/S_k > 1/2$ .

step 1.2step 2.1L7choose

3.2

Therefore $\sum_n (b_n - b_{n+1})$ converges.

step 2.2L5

4.1

So no $N$ witnesses the Cauchy condition for the tolerance $\varepsilon = 1/2$ , and $\sum_n a_n/S_n$ diverges, which is claim 1.

step 3.1L3

5.1

By comparison, $\sum_n \dfrac{a_{n+1}}{S_{n+1}^{2}}$ converges; that series is the $1$ -st tail series of $\sum_n \dfrac{a_n}{S_n^{2}}$ , so the latter converges, which is claim 2.

step 3.2step 2.3L6∎

Remarks

The two claims are not two theorems but one pair of estimates. Divergence comes from bounding a block below by $1 - S_m/S_n$ , which the Cauchy criterion turns into a refutation of convergence; convergence comes from bounding a single term above by a telescoping difference. The first estimate needs $S_n$ to grow without bound and the second needs it to be nondecreasing, and both facts come from divergence of $\sum a_k$ together with positivity of its terms.
The exponent $2$ is not optimal, and this page does not pursue that. The classical refinement replaces $S_n^{2}$ by $S_n^{1+\delta}$ for a rational $\delta > 0$ ; the argument is the same in outline but needs an estimate for $S_n^{-\delta} - S_{n+1}^{-\delta}$ that the tools on this page do not supply cleanly. The square is what claim 1 needs a companion for, and it is enough for every use made of the theorem here.
Positivity is used at every step. It gives $S_n > 0$ , so the quotients exist; it makes $(S_n)$ nondecreasing, which both estimates use; and it makes the terms of the two derived series nonnegative, which is what lets comparison and the boundedness criterion apply to them.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 102 results over 27 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

Divergent series (Wikipedia) (standard reference, not scraped)
K. Knopp, Theory and Application of Infinite Series, Ch. IX (standard reference, not scraped)
Abel-Dini-Pringsheim theorem (Wikipedia) (standard reference, not scraped)

For a divergent series of positive terms with partial sums sks_ksk​, the series ∑ak/sk\sum a_k/s_k∑ak​/sk​ diverges and ∑ak/sk2\sum a_k/s_k^2∑ak​/sk2​ converges