TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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The p-series for a real exponent p converges exactly when p is greater than one

Statement

For every real $p$ , $\sum_{k \geq 1} \frac{1}{k^{p}} converges ⟺ p > 1.$

Facts & Assumptions

Given: A real exponent $p$ .

[L1]

The integral test applies to a nonnegative nonincreasing function on $[0, \infty)$ and compares convergence with boundedness of its proper-integral sequence (The integral test: for $f \geq 0$ nonincreasing on $[0, \infty)$ , $\sum_{k} f (k)$ converges if and only if the sequence $(\int_{0}^{N} f)_{N}$ is bounded, with $\int_{0}^{N} f \leq \sum_{k < N} f (k) \leq f (0) + \int_{0}^{N} f$ ).

[L2]

Positive-base real powers are continuous and differentiable, with the stated power laws (Continuity and derivatives of positive-base real powers, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents).

[L3]

$\int_{1}^{x} d t / t = \log x$ , and $\log x \to + \infty$ as $x \to + \infty$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ ).

[L4]

A convergent series has terms tending to zero (If a series converges then its terms tend to $0$ ).

Proof

technique · direct

1.1

If $p \leq 0$ , then $1 / k^{p} \geq 1$ for $k \geq 1$ , so the terms do not tend to zero and the series diverges.

L2L4

1.2

Suppose $p > 0$ and set $f (t) = (t + 1)^{- p}$ . By [L2] this is nonnegative and nonincreasing on $[0, \infty)$ , and its sampled series is $\sum_{k \geq 0} 1 / (k + 1)^{p}$ .

L1L2

2.1

If $p = 1$ , then $\int_{0}^{N} f (t) d t = \log (N + 1)$ , which is unbounded by [L3].

step 1.2L3

2.2

If $p \neq 1$ , the power derivative gives $\int_{0}^{N} f (t) d t = ((N + 1)^{1 - p} - 1) / (1 - p)$ ; this is bounded exactly when $p > 1$ , using the exponential limits in [L3].

step 1.2L2L3

3.1

The integral test gives convergence exactly for $p > 1$ when $p > 0$ , and step 1.1 handles $p \leq 0$ .

step 1.1step 2.1step 2.2L1∎

Depends on

Used by

Equal radii do not determine convergence on the boundary circle Counterexample
Summable untruncated variances are not necessary Counterexample
The log(1+x) power series diverges at x=-1 Counterexample
Dirac comb Definition
A power series with reciprocal-square coefficients converges uniformly on the closed unit disc Example
Almost-sure conditional convergence Example
k⁻ᵃ membership in ℓᵖ Example
Nonidentical strong law under summable normalized variances Example
Rademacher-series threshold Example
The harmonic complex power series diverges at 1 and converges conditionally at every other point of the unit circle Example
The three series impose separate conditions Example
Stirling's factorial asymptotic holds up to a positive constant Lemma
Finite variance logarithmic rate for iid sums Theorem
Mertens' first theorem for primes Theorem
Polynomial growth functions define tempered distributions Theorem
Strong law under summable normalized variances Theorem

Dependency tree · two levels

45 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)