On An Additive Analogue Of The Function S, By Jozsef Sandor

On an additive analogue of the function S J´ozsef S´andor Babe¸s-Bolyai University, 3400 Cluj-Napoca, Romania The function S, and its dual S∗ are defined by S(n) = min{m ∈ N : n|m!}; S∗ (n) = max{m ∈ N : m!|n} (see e.g. [1]) We now define the following ”additive analogue”, which is defined on a subset of real numbers. Let S(x) = min{m ∈ N : x ≤ m!},

x ∈ (1, ∞)

(1)

S∗ (x) = max{m ∈ N : m! ≤ x},

x ∈ [1, ∞).

(2)

as well as, its dual

Clearly, S(x) = m if x ∈ ((m − 1)!, m!] for m ≥ 2 (for m = 1 it is not defined, as 0! = 1! = 1!), therefore this function is defined for x > 1. In the same manner, S∗ (x) = m if x ∈ [m!, (m + 1)!) for m ≥ 1, i.e. S∗ : [1, ∞) → N (while S : (1, ∞) → N). It is immediate that S(x) =

   S∗ (x) + 1, if x ∈ (k!, (k + 1)!) (k ≥ 1)   S∗ (x),

if x = (k + 1)!

1

(k ≥ 1)

(3)

Therefore, S∗ (x) + 1 ≥ S(x) ≥ S∗ (x), and it will be sufficient to study the function S∗ (x). The following simple properties of S∗ are immediate: 1◦ S∗ is surjective and an increasing function 2◦ S∗ is continuous for all x ∈ [1, ∞)\A, where A = {k!, k ≥ 2}, and since lim S∗ (x) = x%k!

k − 1, lim S∗ (x) = k (k ≥ 2), S∗ is continuous from the right in x = k! (k ≥ 2), but it is x&k!

not continuous from the left. S∗ (x) − S∗ (k!) = 0, it has a rightx&k! x − k!

3◦ S∗ is differentiable on (1, ∞) \ A, and since lim derivative in A ∪ {1}.

4◦ S∗ is Riemann integrable in [a, b] ⊂ R for all a < b. a) If [a, b] ⊂ [k!, (k + 1)!) (k ≥ 1), then clearly Z

b

S∗ (x)dx = k(b − a)

(4)

a

b) On the other hand, since l!

Z

=

k!

(k+1)!

Z

+

k!

Z

(k+2)!

+... +

(k+1)!

(k+l−k)!

Z

(k+l−k−1)!

(where l > k are positive integers), and by Z

(k+1)!

S∗ (x)dx = k[(k + 1)! − k!] = k 2 · k!,

(5)

k!

we get Z

l!

S∗ (x)dx = k 2 · k! + (k + 1)2 (k + 1)! + . . . + [k + (l − k − 1)]2 [k + (l − k − 1)!]

k!

c) Now, if a ∈ [k!, (k + 1)!], b ∈ [l!, (l + 1)!), by Z a

b

=

Z

(k+1)!

+

a

Z

l!

(k+1)!

2

+

Z l!

k

(6)

and (4), (5), (6), we get: b

Z

S∗ (x)dx = k[(k + 1)! − a] + (k + 1)2 (k + 1)! + . . . +

a

+[k + 1 + (l − k − 2)]2 [k + 1 + (l − k − 2)!] + l(b − l!)

(7)

We now prove the following Theorem 1. S∗ (x) ∼

log x log log x

(x → ∞)

(8)

Proof. We need the following Lemma. Let xn > 0, yn > 0,

xn → a > 0 (finite) as n → ∞, where xn , yn → ∞ yn

(n → ∞). Then log xn →1 log yn

Proof. log

(n → ∞).

(9)

xn → log a, i.e. log xn − log y = log a + ε(n), with ε(n) → 0 (n → ∞). So yn log xn log a ε(n) −1= + → 0 + 0 · 0 = 0. log yn log yn log yn

Lemma 2. a)

n log log n! → 1; log n!

log n! → 1; log(n + 1)! log log n! c) → 1 as n → ∞ (10) log log(n + 1)! Proof. a) Since n! ∼ Ce−n nn+1/2 (Stirling’s formula), clearly log n! ∼ n log n, so b) log n n follows by ∼ 1 ((9), since ∼ 1). Now c) is a consequence of b) by the log(n + 1) n+1 Lemma. Again by the Lemma, and log n! ∼ n log n we get b)

log log n! ∼ log(n log n) = log n + log log n ∼ log n and a) follows.

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Now, from the proof of (8), remark that n log log n! S∗ (x) log log x n log log(n + 1)! < < log(n + 1)! log x log n! and the result follows by (10). ∞ X Theorem 2. The series n=1

1 is convergent for α > 1 and divergent for n(S∗ (n))α

a ≤ 1. Proof. By Theorem 1, A

log n log n < S∗ (n) < B log log n log log n

(A, B > 0) for n ≥ n0 > 1, therefore it will be sufficient to study the convergence of ∞ X (log log n)α . n(log n)α n≥n0 The function f (x) = (log log x)α /x(log x)α has a derivative given by x2 (log x)2α f 0 (x) = (log log x)α−1 (log x)α−1 [1 − (log log x)(log x + α)] implying that f 0 (x) < 0 for all sufficiently large x and all α ∈ R. Thus f is strictly X decreasing for x ≥ x0 . By the Cauchy condensation criterion ([2]) we know that an ↔ X 2n a2n (where ↔ means that the two series have the same type of convergence) for (an ) strictly decreasing, an > 0. Now, with an = (log log n)α /n(log n)α we have to study X 2n (log log 2n )α X  log n + a α ↔ , where a, b are constants (a = log log 2, b = 2n (log 2n )α n+b  α log n + a log 2). Arguing as above, (bn ) defined by bn = is a strictly positive, strictly n+b decreasing sequence, so again by Cauchy’s criterion X

n≥m0

bn ↔

X 2n (log 2n + a)α X 2n (nb + a)α X = = cn . (2n + b)α (2n + b)α n≥m n≥m n≥m 0

0

0

cn+1 1 = α−1 , by an easy computation, so D’Alembert’s criterion proves n→∞ cn 2 X 2n (nb + a) , which is clearly the theorem for α 6= 1. But for α = 1 we get the series 2n + b divergent. Now, lim

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References [1] J. S´andor, On certain generalizations of the Smarandache function, Notes Numb. Th. Discr. Math. 5(1999), No.2, 41-51. [2] W. Rudin, Principles of Mathematical Analysis, Second ed., Mc Graw-Hill Company, New York, 1964.

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