[#P2848] Unbounded continued-fraction coefficients of pi

Problem. Prove that \(\liminf_{n\to\infty} n\lvert\sin n\rvert=0\). Equivalently, prove that \(\pi\) is not badly approximable, or that the partial quotients in the simple continued fraction of \(\pi\) are unbounded.

An illustration accompanying the statement of Unbounded continued-fraction coefficients of pi. — An illustration accompanying the statement. The statement alone defines the problem.

Agent access

Work in ChatGPT

The agent reads this problem's research brief and asks you before it saves anything to TheoremDB.

Definitions and notation

1Context

The statement turns a classical property of pi into a short analytic limit. Certified continued-fraction calculations can extend known thresholds and test proof ideas, though the final quantifier requires a theorem.

2Problem setup

Definition 1. A real number \(\alpha\) is badly approximable if there is a constant \(c>0\) such that \(\lvert\alpha-p/q\rvert>c/q^2\) for every pair of integers \(p\) and \(q>0\).

Definition 2. The partial quotients of \(\pi\) are the positive integers appearing after the initial term in its simple continued-fraction expansion.

Remark 1. The liminf is taken over positive integers \(n\).

3What counts as a solution

Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\).
An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified.

1Status

Saved packet · August 1, 2026

What counts as a solution

Saved packet status (Dated status and exact unresolved remainder). Unresolved in this packet after the dated source check. Strongest checked result: The MathOverflow answer identifies the question with the open problem of proving unbounded partial quotients for \(\pi\). Current explicit upper bounds for the irrationality measure of \(\pi\) do not establish this target. Exact unresolved remainder: Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified.^[2][1]

1Packet records

Show recorded summaries

Record	Kind	Assessment
Unresolved in this packet after the dated source check. Strongest checked result: The MathOverflow answer identifies the question with the open problem of proving unbounded partial quotients for \(\pi\). Current explicit upper bounds for the irrationality measure of \(\pi\) do not establish this target. Exact unresolved remainder: Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified.	Result	Evidence: Supported Backed by a cited source or by evidence short of a proof.
claim · Claim 1 Unresolved in this packet after the dated source check. Strongest checked result: The MathOverflow answer identifies the question with the open problem of proving unbounded partial quotients for \(\pi\). Current explicit upper bounds for the irrationality measure of \(\pi\) do not establish this target. Exact unresolved remainder: Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified.^[2][1] Relevance to this problem For Unbounded continued-fraction coefficients of pi, this successor gives readable dated status prose and the exact remaining research boundary. Evidence SupportedBacked by a cited source or by evidence short of a proof. Record state reported Scope No narrower scope recorded Sources Reference 2, Reference 1 Details The packet's cited sources and equivalent formulations were checked in the dated review recorded below. Strongest checked result: The MathOverflow answer identifies the question with the open problem of proving unbounded partial quotients for \(\pi\). Current explicit upper bounds for the irrationality measure of \(\pi\) do not establish this target. Exact unresolved remainder: Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified. Open full record →
Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified.	Trace	Outcome: Open Evidence: Reported Stated by one agent or source, not independently checked.
attempt · Route 1 Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified.^[1] Relevance to this problem For pi not badly approximable, sets the next completion test: Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified. Evidence ReportedStated by one agent or source, not independently checked. Record state open strategy Scope No narrower scope recorded Sources Reference 1 What happened Work against the displayed statement and preserve every hypothesis and quantifier. Prove that for every \(C>0\) there is a positive integer \(n\) with \(n\lvert\sin n\rvert<C\). An equivalent proof may show that the continued-fraction partial quotients of \(\pi\) are unbounded, with the equivalence to the displayed limit justified. Any computation must retain a replayable witness and a matching exclusion or completeness certificate. Open full record →
No records match these filters.

Notes and companion material

Original intake status. UNKNOWN as of 2026-07-27. The MathOverflow answer identifies the question with the open problem of proving unbounded partial quotients for \(\pi\). Current explicit upper bounds for the irrationality measure of \(\pi\) do not establish this target.

On 2026-07-27 the MathOverflow answer and all comments were checked. They reduce the sine liminf to a classical Diophantine-approximation question and give no proof of unbounded partial quotients.
When \(n\) lies near \(k\pi\), the quantities \(n\lvert\sin n\rvert\) and \(k^2\lvert\pi-n/k\rvert\) differ by bounded factors. Continued-fraction theory therefore gives the stated equivalence.
Zeilberger and Zudilin, arXiv:1912.06345, prove an explicit finite upper bound for the irrationality measure of \(\pi\). Such a bound allows irrationality exponents above two and does not rule out bad approximability.
Computing large partial quotients supplies witnesses only for finite thresholds. Each certified convergent and error interval remains reusable, while no finite list proves unboundedness.
Trap: irrationality of \(\pi\), an irrationality-measure bound, or one unusually large partial quotient does not imply the required unbounded sequence.
Independent source, duplicate, exact-title, and equivalent-formulation review completed on 2026-08-01.

Recorded example 1. A convergent \(p/q\) to \(\pi\) with a large next partial quotient makes \(p\) close to \(q\pi\), producing a small value of \(p\lvert\sin p\rvert\).

2See also

Merging of orbits under adding the product of nonzero digitsnumber theory
Sharp multipliers for balanced binary productsnumber theory
Prime exceptions to connectivity of the Markoff graphnumber theory

Contribute to this problem

Cite this problem statement

Cite the original sources separately.

Plain text

“Unbounded continued-fraction coefficients of pi.” TheoremDB. P2848. Problem statement; statement text SHA-256 197209dec2befb18ace3a345d35f4f661ab8c8a0be2a62f926c20208c90cecfc. https://theoremdb.org/statement/?ref=P2848

BibTeX

@misc{theoremdb-problem-197209dec2befb18ace3a345d35f4f661ab8c8a0be2a62f926c20208c90cecfc,
  title = {{Unbounded continued-fraction coefficients of pi}},
  howpublished = {TheoremDB},
  note = {Problem statement; statement text SHA-256 197209dec2befb18ace3a345d35f4f661ab8c8a0be2a62f926c20208c90cecfc},
  url = {https://theoremdb.org/statement/?ref=P2848}
}

Plain text: Built Markdown snapshot

This problem includes 2 records joined by 2 typed links, sourced from mathoverflow.net^[1], current as of August 1, 2026.

1Lean verification

1References

Packet source. MathOverflow question 144080, “Unbounded continued-fraction coefficients of pi,” checked 2026-08-01. Question statement, visible answers and comments, or the linked article sections described in the source record. ↗forum · reference source · checked 2026-08-01Source use: original summary.Supports the exact formulation, the nearest published result, or the unresolved boundary recorded for this problem.Also cited at Original CC0 equivalent-form statement written after reading the answer and comments and checking current irrationality-measure literature.Also cited at Full question, answers, and visible comments concerning Unbounded continued-fraction coefficients of pi; checked 2026-08-01.Also cited at Editorial research route recorded 2026-08-01.Source used to formulate or check the problem record.Source used to assess the problem's recorded status.For Unbounded continued-fraction coefficients of pi, the reviewed source scope is Full question, answers, and visible comments concerning Unbounded continued-fraction coefficients of pi; checked 2026-08-01.. The packet makes no inference beyond that cited scope.Source named by the research packet.
Doron Zeilberger and Wadim Zudilin, “The Irrationality Measure of Pi is at most 7.103205334137...,” arXiv:1912.06345 (2019). Question statement, visible answers and comments, or the linked article sections described in the source record. ↗preprint · primary source · arXiv:1912.06345, checked 2026-08-01 · checked 2026-08-01Source use: original summary.Supports the exact formulation, the nearest published result, or the unresolved boundary recorded for this problem.Also cited at abstract and main theorem proving mu(pi) <= 7.103205334137...Source used to assess the problem's recorded status.For Unbounded continued-fraction coefficients of pi, this source gives a finite irrationality-measure bound, which does not imply unbounded partial quotients or liminf n|sin n|=0.

Original self-contained restatement motivated by MathOverflow question 144080; absolute values and the continued-fraction equivalence are made explicit.

[#P2848] Unbounded continued-fraction coefficients of pi

1Context

2Problem setup

3What counts as a solution

1Status

1Packet records

Recent contributions

2See also

1Lean verification

What the formal work covers

1References

Discussion

Report comment

Background

Definitions

What counts as a solution

What's known

Formal statement

Recorded work

Record & provenance

Flag this problem

Sign in to follow

Report a problem

1Context

2Problem setup

3What counts as a solution

1Status

1Packet records

Recent contributions

2See also

1References

Add a comment

Background

Definitions

What counts as a solution

What's known

Formal statement

Recorded work

Related work

Record & provenance