Factorial digits of π βπ +2
The small perturbation from the reciprocal-factorial series can also
be isolated term by term. Subtracting π β2 leaves the absolutely convergent
correction πΆ =βπβ₯2(π!(π! β1))β1.
The identity 1/(π! β1) =1/π! +1/(π!(π! β1)) gives
π=πΆ+πβ2.(17)
The canonical factorial digits of a real π₯ are ππ(π₯) =βπ!π₯β βπβ(π β1)!π₯β β{0,β¦,π β1} for π β₯2. At a rational endpoint, the floor
convention selects the terminating representation. For example, 1/2 =1/2! also equals βπβ₯3(π β1)/π!, but its canonical
digits are π2 =1 and ππ =0 for π β₯3, not an eventually maximal tail.
With this convention, π₯ is rational
exactly when its digits vanish from some index on. The criterion goes
back to Cantor [5];
Galambos treats the rationality of Cantor series in [6], and Koepf and
Schmersau prove the irrationality direction for nonterminating factorial
expansions whose digits are not eventually maximal [9].
Proposition 4.2 (rationality and factorial residues).
πβββΊβπ!πΆββ‘β2(modπ)for all sufficiently largeΒ π.
Proof. Put π½π =βππ=0π!/π!. Then π½π is an integer, every summand with
π <π is divisible by π and the summand at π =π is 1, so π½π β‘1(modπ); and 0 <π! π βπ½π <1 for π β₯2.
Suppose π is rational. For all
large π both (π β1)! π and π! π are integers, so π β£π! π. Multiplying (17)
by π! and using β βπ! πβ = βπ½π β1 gives βπ!πΆβ =π! π +2 π! βπ½π β1 β‘ β2(modπ).
Conversely, assume the congruence from some index on. Because 0 β€ππ(πΆ) <π, for π β₯3 the congruence is equivalent to
ππ(πΆ) =π β2. Choose π beyond the exceptional indices. The
canonical expansion gives
πΆ=βπ!πΆβπ!+βπ>ππβ2π!,
and the telescope
βπ>π(π β1)/π! =1/π! turns
this into
π=πΆ+πβ2=βπ!πΆβ+1π!+πβπ=21π!,
which is rational.Β β»
The proof applies to every real πΆ, not just this series: eventual digits
ππ(πΆ) =π β2 characterize rationality
of πΆ +π β2. Thus πΆ =2 βπ satisfies the eventual pattern,
whereas πΆ =0 does not. For our
particular positive πΆ, neither
possibility is decided by a finite digit list.
Negating the eventual congruence gives the equivalent formulation
πβββΊ(βπ΅)(βπ>π΅)Β (βπ!πΆβ+2)modπβ 0.(18)
It remains to prove failures of this congruence
at arbitrarily large indices for this particular πΆ. A finite list of failures does not
settle this question, and the equivalence gives no distribution
estimate.
Recovering the finite carries.
To compare canonical digits with the carries, we must account for the
tail removed before taking a floor. Set
πΏπ=π!βπ>π1π!(π!β1),ππ=π{{π!πΆ}<πΏπ}.
Here {π₯} =π₯ ββπ₯β. The
factorial-tail bound gives 0 <πΏπ <1/((π +1)! β1) <1.
Subtracting πΏπ from π!πΆ lowers its floor by one exactly when
{π!πΆ} <πΏπ. Therefore
ππ=π!πβπ=21π!+βπ!πΆβ+1βππ.
Substitution in (12) gives
ππ=πβ1βππ(πΆ)+ππβπππβ1(πβ₯3).(19)
When {π!πΆ} =πΏπ, the scaled partial sum
is integral and ππ =0. For a
smaller fractional part, subtracting even this small tail crosses an
integer. Tail size alone therefore permits neither correction term to be
omitted.