by rpenner » Fri Jun 26, 2015 5:16 pm
Thank you, NameAlreadyTaken2 and Gibborim
The sum over reciprocals of all positive integers except those with a digit of b-1 when expressed in base b, converges.
I don't think Mathematica can prove that
Sum[ If[MemberQ[IntegerDigits[n, b], b-1], 1/n, 0 ], {n, 1, Infinity}]
= Sum[1/FromDigits[IntegerDigits[n, b-1], b], {n, 1, Infinity}]
is an equality or that it converges.
But I think:
FromDigits[IntegerDigits[n, b-1], b]
= n + Sum[ b^k Floor[ n / (b-1)^(1+k) ], {k, 0, Infinity} ]
= n + Sum[ b^k Floor[ n / (b-1)^(1+k) ], {k, 0, Floor[ Log[n] / Log[b-1] ] } ]
≈ n + n Sum[ b^k / (b-1)^(1+k) ], {k, 0, Floor[ Log[n] / Log[b-1] ] } ]
= n ( (b-1) / b )^( -1 - Floor[ Log[n] / Log[b-1] ] )
≈ n ( (b-1) / b )^( -1 - Log[n] / Log[b-1] )
= b/(b-1) * n^(Log[b]/Log[b-1])
which produces (b-1) Zeta[ Log[b]/Log[b-1] ] / b as an estimate ( lower bound ) of the limit.
NameAlreadyTaken2 's estimate is Zeta[1 - Log[(b-1)/b]/Log[b]] = Zeta[ 2 - Log[b-1] / Log[b] ] and when 3 < b < 10, this estimate is between 7/4 and 13/4 larger than my lower bound
(2/3) Zeta[ Log[3] / Log[2] ] ≈ 1.55
Zeta[ 2 - Log[2] / Log[3] ] ≈ 3.31
Sum[1/FromDigits[IntegerDigits[n, 2], 3], {n, 1, 10}] ≈ 2.06
Sum[1/FromDigits[IntegerDigits[n, 2], 3], {n, 1, 100}] ≈ 2.31
Sum[1/FromDigits[IntegerDigits[n, 2], 3], {n, 1, 1000}] ≈ 2.64
Sum[1/FromDigits[IntegerDigits[n, 2], 3], {n, 1, 10000}] ≈ 2.67
(9/10) Zeta[ Log[10] / Log[9] ] ≈ 19.29
Zeta[ 2 - Log[9] / Log[10] ] ≈ 22.43
Sum[1/FromDigits[IntegerDigits[n, 9], 10], {n, 1, 10}] ≈ 2.91
Sum[1/FromDigits[IntegerDigits[n, 9], 10], {n, 1, 100}] ≈ 4.95
Sum[1/FromDigits[IntegerDigits[n, 9], 10], {n, 1, 1000}] ≈ 6.83
Sum[1/FromDigits[IntegerDigits[n, 9], 10], {n, 1, 10000}] ≈ 8.51