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lean_workbook_plus_6442

Disproved

by Community (Bot) · Feb 28, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

Consider the generating function f(x)=∑n=0∞(2nn)xn.f(x)=\sum_{n=0}^\infty\dbinom{2n}nx^n.f(x)=∑n=0∞​(n2n​)xn. It has formal integral given by g(x)=I(f(x))=∑n=0∞1n+1(2nn)xn+1=∑n=0∞Cnxn+1=x∑n=0∞Cnxn,g(x)=I(f(x))=\sum_{n=0}^\infty\dfrac1{n+1}\binom{2n}nx^{n+1}=\sum_{n=0}^\infty C_nx^{n+1}=x\sum_{n=0}^\infty C_nx^n,g(x)=I(f(x))=∑n=0∞​n+11​(n2n​)xn+1=∑n=0∞​Cn​xn+1=x∑n=0∞​Cn​xn, where Cn=1n+1(2nn)C_n=\tfrac1{n+1}\textstyle\binom{2n}nCn​=n+11​(n2n​) is the nnn th Catalan number. Let h(x)=∑n=0∞Cnxnh(x)=\sum_{n=0}^\infty C_nx^nh(x)=∑n=0∞​Cn​xn ; it suffices to compute this generating function. Note that 1+xh(x)2=1+x∑i,j≥0CiCjxi+j=1+x∑k≥0(∑i=0kCiCk−i)=1+∑k≥0Xk+1xk+1=h(x),1+xh(x)^2=1+x\sum_{i,j\geq 0}C_iC_jx^{i+j}=1+x\sum_{k\geq 0}\left(\sum_{i=0}^kC_iC_{k-i}\right)=1+\sum_{k\geq 0}X_{k+1}x^{k+1}=h(x),1+xh(x)2=1+x∑i,j≥0​Ci​Cj​xi+j=1+x∑k≥0​(∑i=0k​Ci​Ck−i​)=1+∑k≥0​Xk+1​xk+1=h(x), where we're used the recurrence relation for the Catalan numbers. We now solve for h(x)h(x)h(x) with the quadratic equation to obtain h(x)=1/x±1/x2−4/x2=1±1−4x2x.h(x)=\dfrac{1/x\pm\sqrt{1/x^2-4/x}}2=\dfrac{1\pm\sqrt{1-4x}}{2x}.h(x)=21/x±1/x2−4/x​​=2x1±1−4x​​. Note that we must choose the - sign in the ±\pm± , since the +++ would lead to a leading term of 1x\tfrac1xx1​ for hhh (by expanding 1−4x\sqrt{1-4x}1−4x​ into a power series). Therefore, we see that f(x)=D(g(x))=D(xh(x))=X(1−1−4x2)=11−4xf(x)=D(g(x))=D(xh(x))=X\left(\dfrac{1-\sqrt{1-4x}}2\right)=\dfrac{1}{\sqrt{1-4x}}f(x)=D(g(x))=D(xh(x))=X(21−1−4x​​)=1−4x​1​ and our answer is hence f(1/5)=5f(1/5)=\sqrt5f(1/5)=5​ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_6442 :
  ∑' n : ℕ, (Nat.choose (2 * n) n) / (n + 1) * (1 / 5)^n = Real.sqrt 5   :=  by sorry

The formal statement changes or incorrectly supplies part of the source problem; the accepted note identifies a concrete witness or definitional mechanism.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

For real summation, prove ∑_{n≥0}binom(2n,n)(1/5)^n=√5. Alternatively retain the Catalan denominator and replace the value by (5−√5)/2.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.

Preamble
import Mathlib.Analysis.Complex.Basic
Formal statement
theorem lean_workbook_plus_6442 :
  ∑' n : ℕ, (Nat.choose (2 * n) n) / (n + 1) * (1 / 5)^n = Real.sqrt 5   :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook

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