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

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