TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$

Statement

Let $p \in Q$ with $p > 0$ . For a natural number $k \geq 1$ write $ι (k) = k \cdot 1_{R}$ for the canonical natural, which is positive (Canonical naturals are positive and strictly increasing), and write $k^{p} : = ι (k)^{p}$ for its rational power (Rational powers $a^{r}$ of a positive base). Then

$\sum_{k \geq 1} \frac{1}{k^{p}} converges ⟺ p > 1.$

In particular the harmonic series $\sum_{k \geq 1} 1 / k$ diverges, at $p = 1$ , and $\sum_{k \geq 1} 1 / k^{2}$ converges, at $p = 2$ .

The index range is not cosmetic. The series starts at $k = 1$ because $1 / 0^{p}$ is undefined: Rational powers $a^{r}$ of a positive base gives $0^{p} = 0$ for rational $p > 0$ , and $0$ has no inverse. Sequences here are functions on $N$ and $N$ contains $0$ (Series, partial sums, convergence and the sum, divergence, and the tail series), so the object named above is a series from the starting index $1$ in the sense of Series, partial sums, convergence and the sum, divergence, and the tail series, not a series of a sequence on $N$ .

The exponent is rational, and that is a limitation of this page. Rational powers of a positive base are what Rational powers $a^{r}$ of a positive base supplies; real exponents require the exponential and the logarithm, which this library develops later. The statement above is therefore the full $p$ -series theorem for every exponent this page can name.

Facts & Assumptions

Given: A rational $p > 0$ and the family $a_{k} : = 1 / k^{p} = ι (k)^{- p}$ , defined for naturals $k \geq 1$ (Rational powers $a^{r}$ of a positive base, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

Rational powers of a positive base are positive, and $a^{r + s} = a^{r} a^{s}$ , $(a^{r})^{s} = a^{r s}$ , $a^{- r} = 1 / a^{r}$ for $a > 0$ and rationals $r, s$ (Laws of rational exponents).

[L2]

Monotonicity of rational powers: for rational $t > 0$ and $0 < a < b$ one has $a^{t} < b^{t}$ ; and for $a > 1$ and rationals $r < s$ one has $a^{r} < a^{s}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L3]

The integer power and the rational power agree at an integer exponent: for $a > 0$ and $n \in Z$ , $a^{n}$ read as in Integer powers $a^{m}$ equals $a^{n}$ read as in Rational powers $a^{r}$ of a positive base, since $n = n / 1$ and $a^{1 / 1} = a$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base). In particular $a^{0} = 1$ .

[L4]

Reciprocation reverses the order on the positives: $0 < a < b$ implies $0 < 1 / b < 1 / a$ (Inverses of positives are positive, and reciprocation reverses order).

[L5]

Condensation: for a family $(x_{k})_{k \geq 1}$ that is nonnegative and nonincreasing, $\sum_{k \geq 1} x_{k}$ converges if and only if $\sum_{j \geq 0} 2^{j} x_{2^{j}}$ converges (For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L6]

The geometric series: $\sum_{j \geq 0} r^{j}$ converges if and only if $∣ r ∣ < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Basic properties of the absolute value).

[L7]

The canonical naturals are positive and order preserving: $0 < ι (1) \leq ι (j) < ι (k)$ for naturals $1 \leq j < k$ , and $ι (2) = 2 > 1$ (Canonical naturals are positive and strictly increasing).

Proof

technique · direct

1.1givenL7L1

For every natural $k \geq 1$ the base $ι (k)$ is positive, so $a_{k} = ι (k)^{- p}$ is defined and positive; in particular the family is nonnegative.

1.2givenL7L2L4L1

For naturals $1 \leq j < k$ we have $0 < ι (j) < ι (k)$ , hence $ι (j)^{p} < ι (k)^{p}$ since $p > 0$ , hence $a_{j} = 1 / ι (j)^{p} > 1 / ι (k)^{p} = a_{k}$ ; and for $j = k$ the two are equal. So $a_{j} \geq a_{k}$ whenever $1 \leq j \leq k$ .

1.3L1L3L7algebra

For every $j \in N$ the base $2^{j}$ is positive and, reading the exponent $j$ as a rational, $2^{j} a_{2^{j}} = 2^{j} (2^{j})^{- p} = 2^{j} \cdot 2^{- j p} = 2^{j - j p} = 2^{(1 - p) j} = (2^{1 - p})^{j}$ .

1.4L2L3L7

Since $2 > 1$ , the map $t \mapsto 2^{t}$ is strictly increasing on $Q$ and $2^{0} = 1$ ; hence $r = 2^{1 - p} < 1 = 2^{0}$ holds exactly when $1 - p < 0$ , that is exactly when $p > 1$ .

2.1step 1.1step 1.2L5

Condensation applies to $(a_{k})_{k \geq 1}$ : $\sum_{k \geq 1} a_{k}$ converges if and only if $\sum_{j \geq 0} 2^{j} a_{2^{j}}$ converges.

2.2step 1.3L1L3

So the condensed series is the geometric series $\sum_{j \geq 0} r^{j}$ with $r : = 2^{1 - p}$ , and $r > 0$ , so $∣ r ∣ = r$ .

3.1step 2.2L6

By the geometric series theorem, $\sum_{j \geq 0} r^{j}$ converges if and only if $r < 1$ .

4.1step 2.1step 2.2step 3.1step 1.4∎

Chaining the three equivalences: $\sum_{k \geq 1} 1 / k^{p}$ converges $⟺$ the condensed series converges $⟺$ $r < 1$ $⟺$ $p > 1$ .

Remarks

Where the threshold comes from. Condensation turns the $p$ -series into a geometric series of ratio $2^{1 - p}$ , and the geometric threshold $r = 1$ pulls back to $p = 1$ . Nothing about the number $1$ is special to the $p$ -series; it is the exponent at which the condensed terms stop shrinking.
At $p = 1$ the condensed series is $\sum_{j \geq 0} 1$ . Its terms do not tend to $0$ , so it diverges, and with it the harmonic series. That instance is worked out on the companion page, together with the older block argument that does not use condensation at all.
Only rational exponents are covered, and the gap is real. For irrational $p$ the expression $k^{p}$ has no meaning in this library yet, so the statement is not merely unproved there, it is unstatable. The same limitation is what keeps the Bertrand-type series $\sum 1 / (k (\log k)^{p})$ off this page entirely, the logarithm not being available.

For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For rational p>0p>0, ∑1/kp∑1/kp converges iff p>1p>1

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$