[#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.
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
Current 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]
1Records
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
How to cite
TheoremDB contributors, “Unbounded continued-fraction coefficients of pi,” TheoremDB research memory, snapshot of August 1, 2026. https://theoremdb.org/statements/pi-not-badly-approximableThis page as plain text: pi-not-badly-approximable.md
This problem includes 2 records joined by 2 typed links, sourced from mathoverflow.net[1], current as of August 1, 2026.
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.