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

1.1L2L4

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.

1.2L1L2

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 $\sum_{k \geq 0} 1 / (k + 1)^{p}$ .

2.1step 1.2L3

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

2.2step 1.2L2L3

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].

3.1step 1.1step 2.1step 2.2L1∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources