Amer. Math. Monthly, 115 (2008), p. 933-938.
A curious result related to Kempner’s series Bakir FARHI
[email protected] Abstract It is well known since A. J. Kempner’s work that the series of the reciprocals of the positive integers whose the decimal representation does not contain any digit 9, is convergent. This result was extended by F. Irwin and others to deal with the series of the reciprocals of the positive integers whose the decimal representation contains only a limited quantity of each digit of a given nonempty set of digits. Actually, such series are known to be all convergent. Here, letting S (r) (r ∈ N) denote the series of the reciprocal of the positive integers whose the decimal representation contains the digit 9 exactly r times, the impressive obtained result is that S (r) tends to 10 log 10 as r tends to infinity!
MSC: 40A05. Keywords: Kempner’s series; Harmonic series; Series with deleted terms.
1
Introduction
Throughout this article, we let N∗ denote the set N \ {0} of positive integers. We let E (r) (r ∈ N) denote the set of all positive integers whose decimal repre(r) sentation contains the digit 9 exactly r times. We also let En (r, n ∈ N) denote the set E (r) ∩ [10n , 10n+1 [. We clearly have for any r ∈ N: G E (r) = En(r) . n∈N
For all r, n ∈ N, write:
X 1 k (r)
Sn(r) :=
k∈En
and for all r ∈ N, write: S (r) :=
X 1 X = Sn(r) . k (r) n∈N
k∈E
1
In 1914, A. J. Kempner [3] showed that the series S (0) converges. After him, several generalizations were obtained by several authors. Among others, F. Irwin [2] showed that the series derived from the harmonic series 1 + 21 + 13 + · · · by including only those terms whose denominators contain a limited quantity of each digit of a given nonempty set of digits is convergent. It follows in particular that the series S (r) converges for all natural numbers r. In this paper, the main result obtained is that the sequence (S (r) )r≥1 decreases and converges to 10 log 10 (see Theorem 3). As a consequence, we deduce that we have S (r) > 10 log 10 ' 23.025 (∀r ≥ 1). So, according to the calculations of R. Baillie [1], we have the unexpected inequality S (0) ' 22.920 < S (1) . We must notice that the approximate numerical values of the S (r) ’s (r ≥ 1) are very difficult to calculate. In the last section of the paper, we state a generalization of our main result by taking instead of the digit 9 any other digit d ∈ {0, 1, . . . , 9}.
2
The Results (r)
Suppose r, n ∈ N∗ . If a positive integer k belongs to En , then writing k = 10t + ` with t ∈ N and ` ∈ {0, 1, . . . , 9}, we clearly have either (r) • t ∈ En−1 and ` ∈ {0, 1, . . . , 8}, or (r−1) • t ∈ En−1 and ` = 9. It follows that Sn(r)
=
8 X X `=0
(r) t∈En−1
X 1 1 + . 10t + ` 10t + 9 (r−1)
(1)
t∈En−1
(r)
We will find it useful to approximate Sn with the simpler formula Tn(r)
:=
8 X X `=0
(r) t∈En−1
X 1 1 9 (r) 1 (r−1) + = Sn−1 + Sn−1 . 10t 10t 10 10 (r−1)
(2)
t∈En−1
The error in this approximation is given by
So, we have Sn(r) =
Cn,r := Tn(r) − Sn(r) .
(3)
1 (r−1) 9 (r) Sn−1 + Sn−1 − Cn,r . 10 10
(4)
2
This identity will play an important role in what follows. Our first proposition shows that the errors Cn,r are not very large. Proposition 1 The real numbers Cn,r (r, n ∈ N∗ ) are all nonnegative and we have ∞ X ∞ X Cn,r < ∞. r=1 n=1
Proof. From (1) and the first equality of (2), we have for all r, n ∈ N∗ : Cn,r =
8 X X `=1
(r) t∈En−1
X ` 9 + . 10t(10t + `) 10t(10t + 9) (r−1)
(5)
t∈En−1
This last identity shows that we have Cn,r ≥ 0 (∀r, n ∈ N∗ ). Moreover, using (5), we have for all r, n ∈ N∗ : Cn,r ≤
9 X 1 9 X 1 + . 25 t2 100 t2 (r) (r−1) t∈En−1
t∈En−1
(r)
Since the sets En (r, n ∈ N) form a partition of N∗ , it follows that: ∞ X ∞ X
µ Cn,r ≤
r=1 n=1
9 9 + 25 100
¶X ∞ t=1
1 9 π2 = < ∞. t2 20 6
The proof is complete.
¥
For the following, put for all r ∈ N∗ : Cr :=
∞ X
Cn,r .
n=1
According to (5), we clearly have Cn,r > 0 for all n, r ∈ N∗ such that n ≥ r. Consequently, we have Cr > 0 for all r ∈ N∗ . Proposition 2 For all r ∈ N, the series S (r) converges. In addition, we have: S (1) = S (0) − 10C1 +
10 9
(6)
and S (r) = S (r−1) − 10Cr
3
(∀r ≥ 2).
(7)
Proof. The fact that the series S (r) (r ∈ N) are all convergent is already known (see, e.g., [2]). Let us prove the relations (6) and (7) of the proposition. Using the relation (4), we have for all r ∈ N∗ : S
(r)
= =
∞ X
n=1 ∞ µ X n=1
(r)
Sn(r) + S0
1 (r−1) 9 (r) Sn−1 + Sn−1 − Cn,r 10 10
¶ (r)
+ S0
∞ ∞ ∞ 9 X (r) 1 X (r−1) X (r) = + − Cn,r + S0 S S 10 n=1 n−1 10 n=1 n−1 n=1 ∞ ∞ ∞ 9 X (r) 1 X (r−1) X (r) = Sn + Sn − Cn,r + S0 10 n=0 10 n=0 n=1
=
9 (r) 1 (r) S + S (r−1) − Cr + S0 . 10 10
Hence: (r)
S (r) = S (r−1) − 10Cr + 10S0 ( S (r−1) − 10Cr + 10 if r = 1, 9 = S (r−1) − 10Cr if r ≥ 2. This confirms the required formulas (6) and (7) of the proposition and finishes this proof. ¥ We now arrive at the most important and completely new result of this paper: Theorem 3 The sequence (S (r) )r≥1 decreases and converges to 10 log 10. In particular, we have: S (r) > 10 log 10 (∀r ≥ 1). Proof. Since Cr > 0 (∀r ≥ 1), the formula (7) of Proposition 2 shows that the sequence (S (r) )r≥1 decreases. Since this sequence is positive (so bounded from below by 0), it is necessarily convergent. It remains to calculate its limit as r
4
tends to infinity. We have for all integer R ≥ 2: S
(R)
=
R X ¡
¢ S (r) − S (r−1) + S (1)
r=2
=
R X
(−10Cr ) + S (0) − 10C1 +
r=2
= S (0) − 10
R X
Cr +
r=1
10 9
10 . 9
Hence: lim S
(R)
R→∞
(according to (6) and (7))
=S
Now, let us calculate the sum and (3), we have for all r, n ∈ N∗ :
(0)
− 10
∞ X r=1
P∞ r=1
Cr =
10 . 9 P∞
Cr +
P∞ r=1
n=1
(8) Cn,r . From (1), (2),
Cn,r := Tn(r) − Sn(r) ¶ ¶ 8 X X µ 1 X µ 1 1 1 = − + − . 10t 10t + ` 10t 10t + 9 (r) (r−1) `=0 t∈En−1
t∈En−1
(r)
By remarking that the sets En−1 (r, n ∈ N∗ ) form a partition of N∗ \ E (0) and (r−1) that the sets En−1 (r, n ∈ N∗ ) form a partition of N∗ , we deduce that: ¶ X ¶ ∞ X ∞ 8 ∞ µ X X X µ 1 1 1 1 Cn,r = − + − 10t 10t + ` 10t 10t + 9 (0) ∗ r=1 n=1 t=1 `=0 t∈N \E
¶ µ ¶ X ¶ 8 X ∞ µ ∞ µ X X 1 1 1 1 1 1 = − − − + − 10t 10t + ` 10t 10t + ` t=1 10t 10t + 9 (0) t=1 `=0 =
9 X ∞ µ X `=0 t=1
=
9 X ∞ µ X `=0 t=1
Since
1 1 − 10t 10t + ` 1 1 − 10t 10t + `
P
t∈E (0)
t∈E
¶ −
8 X
¶ X µ 1 1 − 10t 10t + ` (0)
`=0 t∈E
¶ −
8 1 9 X 1 X X + . 10 t 10t + ` (0) (0) `=0 t∈E
t∈E
1/t = S (0) and 8 X X `=0 t∈E (0)
8 8 X 1 X X 1 1 1 = − = S (0) − 10t + ` t t t (0) t=1 t=1 t∈E
5
(because {10t + ` | t ∈ E (0) , ` ∈ {0, 1, . . . , 8}} = E (0) \ {1, 2, . . . , 8}), it follows that: 8 ∞ X ∞ X 1 (0) X 1 , (9) Cn,r = ∆ + S − 10 t r=1 n=1 t=1 where ¶ X ∞ 9 X ∞ µ X 1 1 1 = ∆ := − − t 10t 10t + ` t=1 t=1 `=0
X 10t≤m
