TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $p$ , $\sum_{k\ge1}\frac1{k^p}\text{ converges}\quad\Longleftrightarrow\quad p>1.$

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 $f \ge 0$ nonincreasing on $[0,\infty)$ , $\sum_k f(k)$ converges if and only if the sequence $\bigl(\int_0^N f\bigr)_N$ is bounded, with $\int_0^N f \le \sum_{k<N} f(k) \le f(0) + \int_0^N f$ ).

[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]

$\int_1^xdt/t=\log x$ , 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 $p\le0$ , then $1/k^p\ge1$ for $k\ge1$ , 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 $\sum_{k\ge0}1/(k+1)^p$ .

L1L2

2.1

If $p=1$ , then $\int_0^Nf(t)\,dt=\log(N+1)$ , which is unbounded by [L3].

step 1.2L3

2.2

If $p\ne1$ , the power derivative gives $\int_0^Nf(t)\,dt=((N+1)^{1-p}-1)/(1-p)$ ; 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 $p\le0$ .

step 1.1step 2.1step 2.2L1∎

Depends on

Used by

The log(1+x) power series diverges at x=-1 Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 120 results over 19 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

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)