# A Study Of Slow Increasing Functions And Their Applications To Some Sequences Of Integers

DOI : 10.17577/IJERTV2IS60075

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#### A Study Of Slow Increasing Functions And Their Applications To Some Sequences Of Integers

K. Santosh Reddy*

K. Madhusudhan Reddy and B. Chandra Sekhar

ABSTRACT

In this article we first define a slow increasing function. We investigate some basic properties of slow increasing function. In addition, several applications in some some sequences of integers using the theory of slow increasing functions.

KEYWORDS. Slow Increasing Functions, asymptotically equivalent, sequence of positive integers.

1. INTRODUCTION

Slow increasing functions are defined as follows.

1. Definition. Let f : a, 0, be a continuously differentiable function such that

f 0 and lim f (x) . Then f is said to be a slow increasing function (s.i.f. in short) if

x

lim xf (x) 0

x f (x)

Write F f : f

2. Examples. (i)

is a s.i.f..

f (x) log x, x 1is a s.i.f.

Note that lim f (x) lim log x and f (x) 1 ,x 1 and f is continuous

Moreover

x

x

lim

x

xf (x) lim 1

x 0

x

(ii) f (x) loglog x, x e is also a s.i.f.

f (x)

x x

log x

2. SOME PROPERTIES

1. Theorem. Let f , g F

and let 0, c 0 be to constants then we have

i) f c

1. f c

2. cf (iv) fg (v) f (vi) f g

(vii) log f (viii) f g

all lie in F .

Proof: Given that f , g F and 0, c 0 be constants.

Proof of (i), (ii), (iii), and (iv) follows the definition 1.1

1. Let h f

Note that lim h(x) lim f (x) , and h(x) f (x) 1 f (x) 0, and h is continuous

x x

lim xh(x) lim x f (x) 1 f (x) lim xf (x) 0.

h f F

Moreover

x

h(x)

x

f (x)

x

f (x)

Hence

2. Let h f g i.e h(x) f (g(x))

Note that lim h(x) lim f (g(x)) , and h(x) f (g(x))g(x) 0, and h is continuous

x x

Moreover

lim xh(x) lim xf (g(x))g(x) lim g(x) f (g(x)) xg(x) 0. Hence h f g F

x

h(x)

x

f (g(x))

x

f (g(x))

g(x)

3. Let h log f

Note that lim h(x) lim log f (x) , and h(x) f (x) 0, and h is continuous

x

x

f (x)

Moreover

x f (x)

lim xh(x) lim f (x) lim x f (x) 1 0.

Hence h log f F

x

h(x)

x log f (x)

x

f (x) log f (x)

4. Let h f g

For sufficientely large x , we have 0 xf

xf

and 0 xg

xg

f g f f g g

0 lim xh(x) lim xf lim xg 0

By adding the above, we get

x

h(x)

x f

x g

lim xh(x) 0

Hence h f g F

x h(x)

2. Theorem. Let f , g F. Define h(x) f (x ) and k(x) f (x g(x)) for each x, then h,k F.

Proof: Given that

Let

f , g F. Define h(x) f (x ) and k(x) f (x g(x)) for each x.

h(x) f (x )

Note that lim h(x) lim f (x ) , and h(x) f (x ) x 1 0, and h is continuous

x

x

lim xh(x) lim xf (x ) x 1 lim x f (x ) 0

Moreover

x

h(x)

x

f (x )

x

f (x )

Hence

Let k(x) f (x g(x))

h(x) f (x ) is s.i.f.

Note that lim k(x) lim f (x g(x)) , and

x x

k(x) f (x g(x)) x 1g(x) x g(x) 0 and k is continuous

Moreover

lim

x

xk(x)

k(x)

lim

x

xf (x g(x)) x 1g(x) x g(x) f (x g(x))

lim

x g(x) f (x g(x))

lim

x g(x) f (x g(x)) xg(x)

x

f (x g(x))

x

f (x g(x))

g(x)

0

0

Therefore k(x) f (x g(x)) is s.i.f. Hence h, k F

f (x)

d f (x) f

3. Theorem. Let f , g F

be such that lim

x g(x)

and

dx g(x) 0. Then

F.

g

Proof: Given that

f , g F, lim

f (x)

and

d f (x)

0

Let Moreover

h(x)

f (x)

g(x)

x g(x)

and h(x)

dx g(x)

f (x)g(x) f (x)g(x)

g(x)2

x f (x)g(x) f (x)g(x)

xh(x)

g(x)2

xf (x)

xg(x)

lim

lim lim

• lim 0

Hence

x

h(x)

x

f (x)

g(x)

f F g

x

f (x)

x

g(x)

4. Theorem. Let

and lim h(x)

x

h : a, 0,

be a continuously differentiable function such that

h(x) 0

1. Define g(x) h(log x) . Then g F lim h(x) 0

x h(x)

2. Define k(x) eh(x) .Then k F lim xh(x) 0

x

Proof: Given that h (x) 0 and lim h(x)

x

1. Define

g(x) h(log x) then g(x) h(log x)

x

x h(log x)

Suppose g F then g satisfies lim xg (x) 0

i.e. lim x 0

lim h (log x) 0

x

g(x)

x

h(t)

h(log x)

h(x)

x h(log x)

Put t log x so that x t lim 0

t h(t)

i.e.

h(x)

lim

x h(x)

0 .

Conversely suppose

lim 0

x h(x)

Put t ex

so that

x= log t and

x t

lim

h(x)

lim

h(log t) 0

tg(t)

h(log t)

x h(x)

h(x)

t h(log t)

Now

2. Like proof of (i)

lim

t

g(t)

lim

t h(log t)

lim

x h(x)

0.

Hence g F

5. Theorem. If f F

then lim

x

log f (x) log x

0.

f (x)

log f (x)

f (x)

Proof: Given that

f F,

lim

lim

(byLHospitals rule)

x

lim

x

log x

xf '(x) f (x)

x

0

1

x

x

i.e. lim

x

log f (x) log x

0 .

6. Theorem. f F if and only if to each 0 there exists x such that d f (x) 0, x x

dx x

d f (x)

f (x)x f (x) x 1

f (x) xf (x)

Proof: We have

dx x

x2

x 1

f (x)

Suppose f F

then lim

x

xf '(x) 0

f (x)

i.e. For each 0 there exists x such that x x

And

xf '(x) 0 ,

x x

d f (x) 0,

x x

f (x)

dx x

To prove the converse assumes that the condition holds. Let 0 be given. Then there exist x such that x x

We have, by hypothesis

d f (x) 0

this implies that

xf '(x) 0 ,

x x

dx x

xf (x)

xf '(x)

f (x)

i.e.

f (x)

0 as x lim

x

f (x)

0.

Therefore f F.

7. Theorem. If f F then lim f (x) 0 , for all

x x

Proof: For any with 0 , we get by Theorem 2.6,

d f (x) 0,

x x

for some x

This implies that

f (x)

f (x)

x is decreasing for x x

dx x

Hence bounded above, say, by M

x

f (x)

That is, there exists M > 0 such that 0

x

M , x x

lim f (x) lim f (x) 1 0

x x

x x

x

8. Note. We know that each f F is an icreasing function. Moreover by the above theorem it is clear

f (x)

that lim = 0, >0 . This shows that the increasing nature of f is slow. That is f does not increase

x x

rapidly. This justifies the name given to the members of F.

From the above theorem, we have the following results.

9. Theorem. If f F then lim f (x) 0 and lim f (x) 0.

x x

f (x)

x

Proof: In Theorem 2.7 put 1, toget lim = 0 .

x

If f F , then lim

x

x

xf (x) 0

f (x)

Since

lim f (x) =0

we must have lim f (x)=0.

x x

x

10. Theorem. Let f F then for any 1and , the series n f (n)

n1

diverges to .

1

1

1

Proof: We write

n f (n)

n1

n f (n)

n1 n

we know that the series 1 diverges to

n1 n

Given 1 1 0

If 0 then

lim n 1 f (n)

n

n n 1

If 0 then

lim

n f (n)

lim

n

f (n)

(from Theorem 2.7)

n

i.e. n f (n)

n1

diverges to

An important byproduct of the above theorem is the following result.

x

t f (t) dt

11. Theorem. Let f F. Then for any 1 and , lim a 1.

x x 1 f (x)

1

Proof: From Theorem 2.10, we have lim x 1 f (x)

n

x

x

1

lim

x 1

f (x) ,

x

1,

From Theorem 2.10, we have t f (t) lim t f (t) dt

x

t f (t) dt

t 1

x

a

x f (x)

Consider

lim a lim

(byLHospitalss rule)

x x 1 f (x)

x x 1

x f (x)

1

1

1

f (x)

f (x)

x f (x)

lim 1

x xf (x)

x f (x) 1

1 f (x)

12. Definition.Let f , g :a, 0,

1. If lim f (x) 1, then f is is said to asymptotically equivalent to g . We describe this by writing f

x g(x)

• g .

2. f

(g) Means f

Ag for some

A 0. In this case we say that f is of large order g.

3. f

(g) Means lim f (x) 0 . In this case we say that f is of small order

x g(x)

g.

f (x) xn

13. Examples. (i) Consider

f (x) xn , g(x) xn x, for all x 0 and

lim

x g(x)

lim 1

x xn x

Therefore

f g.

1. x (10x) Because

x 1 x 1 (10x).

10x 10 10

2. x 1 (x2 ) Becuase lim x 1 0.

x x2

As a result of the Theorem 2.11, we get the following results as particular cases.

14. Theorem. Let

f F. Then we have the following statements.

x

(i) f (t)

a

dt xf (x)

(ii)

x

f (t)dt xf (x)

a

(iii)

1 dt

x

x

a f (t)

x

f (x)

Proof: Let f F

1. Put = 0 in Theorem 2.11, we get

x

x

x

f (t) dt

lim a x

xf (x)

1 f (t) dt xf (x)

a

2. Put = 0, = 1 in Theorem 2.11, we get

x

f (t)dt x

lim a 1 f (t)dt xf (x)

x xf (x) a

3. Put = 0, = -1 in Theorem 2.11, we get

1

1

x

dt

x 1 x

15. Theorem. Let f F. Then

lim a x

f (t) 1

x

f (x)

dt

a f (t)

f (x)

(i) lim f (x c) 1, For any c (ii) If f (x) is decreasing then lim f (cx) 1, for any c

x

f (x)

x

f (x)

Proof: Let f F

1. Case (a). Suppose c 0

By Lagranges mean value theorem, There exists a t x, x c such that

f x c f (x) (x c x) f (t)

0 f x c f (x) cf (t)

,

,

0 lim f x c f (x) lim cf (t)

f (x)

t x, x c

f (x)

x

f (x)

x

f (x)

lim f x c 1 0 , since lim f (x) 0 (by Theorem 2.9)

x

f (x)

x

lim f x c 1.

Case (b). Suppose c 0

x f (x)

By Lagranges mean value theorem there exists t x c, x such that

f (x) f x c (x x c) f (t)

0 f (x) f x c cf (t)

f (x)

,

,

0 lim f (x) f x c c lim f (t)

f (x)

t x c, x

x

f (x)

x f (x)

lim f x c 1 0 , since lim f (x) 0

(by Theorem 2.9)

x

f (x)

x

lim f x c 1.

2. Case (a). Suppose c 1

x f (x)

By Lagranges mean value theorem there exists t x, cx such that

f cx f (x) (cx x) f (t)

0 f cx f (x) (c 1)xf (t)

f (x)

,

,

0 lim f cx f (x) (c 1) lim xf (t)

f (x)

t x, cx

And

f (x) is decreasing

x

f (x)

f (x) f (t)

x

f (x)

There fore

lim f cx 1 0 , since lim f (x) 0

(by Theorem 2.9)

x

f (x)

x

lim f cx 1.

Case (b). Suppose c 1

x f (x)

By Lagrange mean value theorem there exists t cx, x such that

f (x) f cx (x cx) f (t)

0 f (x) f cx (1 c)xf (t)

f (x)

,

,

0 lim f (x) f cx (1 c) lim xf (t)

f (x)

t cx, x

x

f (x)

x

f (x)

And

f (x) is decreasing

f (x) f (t)

There fore

lim f cx 1 0 , since lim f (x) 0

(by Theorem 2.9)

x

f (x)

x

lim f cx 1.

x f (x)

16. Theorem. Suppose f F is such that f (x) is decreasing. If 0 c1 c2 and g is a function such that

f (g(x)x)

1 2

1 2

c g(x) c then lim

x

Proof: Suppose f F is such that f (x) is decreasing

f (x)

1.

If 0 c1 g(x) c2 f (c1x) f (g(x)x) f (c2 x) since f is decreasing

f (c1x)

f (g(x)x)

f (c2 x)

lim f (c1x) lim f (g(x)x) lim f (c2 x)

f (x)

f (x)

f (x)

x

f (x)

x

f (x)

x

f (x)

1 lim

x

f (g(x)x) 1

f (x)

(By Theorem 2.15)

lim

x

f (g(x)x)

f (x)

1.

3. APPLICATIONS OF SLOW INCREASING FUNCTIONS TO SOME SEQUENCES OF INTEGERS

This topic is aimed at applications in some special sequences of positive integers. Infact several asymptotic results related to these integer sequences are derived by using the theory of Slow Increasing Functions.

We begin with the following important definition.

Let f F . Through out this chapter (an ) denotes astrictly increasing sequence of positive integers such that

a 1 And lim an

1 for some s 1.

(1)

1 n ns f (n)

n

n

There exist several such sequences.

i.e.

a ns f (n)

For example an pn , the sequence of prime numbers in increasing order,

f (x) log x and s 1.

By prime number theorem we have lim pn 1.

n n log n

1. Definition. Let (an ) be asequence as described above. Then for any x 0 , define (x) 1

an x

The number of an that do not exceed x .

2. Theorem. If (an ) satisfies (1) and g F, then

1. a a

2. lim an1 an =0

3. l o g a log a

4. n1 n

n an

n1 n

g(an1) g(an )

5. log an s log n (vi) loglog an loglog n

(x)

(vii) lim =0

Proof: Let (an ) satisfies (1) and g F

x x

a (n 1)s f

(n 1)

1 s

f (n 1)

1. Consider

lim

n1

= lim

li m 1

lim =1 By

Theorem 2.15

n an

n

ns f (n)

n

n n

f (n)

2. We have

a a

an1 an

lim an1 1 lim an1 1 0 lim an1 an 0

n1 n

n an

n an

n an

3. Consider

lim an1 1 log lim an1 log1 lim log an1 0

n an

n an

n

an

limlog an1

• log an

0 lim log an1 log an 0

n

n

log an

lim log an1 1

i.e. l o g an1

• log an

n

log an

4. As a a , g F, we have

lim g(an1 ) 1

g(a ) g(a )

n1 n

n g(an )

n1 n

5. We have

an

ns f (n)

log an

log ns f (n)

log an

slog n log f (n)

log an

1 log f (n)

lim log an = 1 1 lim log f (n)

lim log an 1 By

slog n

Theorem 2.5

slog n

n slog n s n

log n

n s log n

6. We have logan s log n

i.e. logan s log n

loglogan logs log n

loglogan log s loglog n

loglogan

log s 1

lim loglogan log s lim 1 1

log log n

log log n

n log log n

n log log n

lim loglogan 1

n log log n

i.e. loglogan

• loglog n.

7. We have

(x) 1

an x

Let n be the largest index such that a

x then

(x) n

(x) n0 n0

0 n0

0 x x a

n

n

0

0 lim (x) lim n0

0 lim (x) 0

lim (x) 0

x x

n0 a

n

n

0

x x

x x

3. Theorem. Suppose an

g(a ) lg(n) g( (x)) 1 g(x).

satisfies (1) and g F,

then for l 1,

n

Proof: First suppose that

l

g( (x)) 1 g(x) g( (a )) 1 g(a )

l n l n

g(n) 1 g(a ) g(a ) lg(n) . ( (a ) n )

l n n n

Conversely suppose

g(a ) lg(n) g(a ) lg( (a ))

lim g( (an )) 1

n n n

n 1 g(a )

l n

(2)

If an x an1 we have

(an ) x an1

g( (an )) g( (x)) g( (an1))

since

And

g F.

g(a ) g x g a

1 g(a ) 1 g x 1 g a

(l 1)

n n1

l n l l n1

lim g( (an )) lim g( (x)) lim g( (an ))

n

1 g(a )

x

1 g(x)

n

1 g(a )

l n l l n

(a

a )

1 lim g( (x)) 1 by 2

n1 n

x

1 g x l

lim g( (x)) 1

g( (x)) 1 g(x).

x

1 g x l

l

4. Theorem. Suppose an

satisfies (1) and g F , then (i)

log an

• s log n log (x) 1 log x

s

s

(ii) loglog an log logn log log (x) log log(x).

Proof: Given an satisfies (1) and g F

1. In Theorem 3.3 put g(an ) log an , g(n) log n and l s . And we have logan s log n.

2. In Theorem 3.3 put g(an ) loglog an , g(n) loglog n and l s

And we have loglogan loglog n.

bn

bn

We make use the following well known result in proving theorems to follow.

5. Result. Let

bn

and

cn be two series of positive terms such that lim

1. If

cn is divergent

n1

n1

n cn

n1

n

n

bk

n

n

then it is known that lim k 1 1.

n

ck

k 1

6. Theorem. Let f F, an satisfies (1) and f (an )

f (n). Then

k

k

x x 1

an nf (n) (x)

(x) dt f (a ) x.

f (x) f (t)

Proof: Given an

(3)

satisfies (1) and f F

a

and

ak x

f (an )

f (n)

First suppose that

(x) x

f (x)

(an )

an

f (an )

n an

f (an )

an nf (an )

an nf (n)

By (3)

Conversely suppose

a nf (n)

a (a ) f (n)

(a ) an

n

(a )

n n

an by (3)

n

lim (an ) 1

f (n)

n

(4)

f (an )

n an

f (an )

If an x an1 we have

(an ) x an1

We have by Theorem 2.6

f (x) x

has negative derivative

f (x) x

is decreasing

x

f (x)

is increasing

an

x

an1

lim (an ) lim (x) lim (an )

(a a )

f (a )

f (x)

f (a )

n an x x n an

n1 n

n n1

f (an )

f (x)

f (an )

1 lim x 1

By (4)

x

lim 1

(x) x

x x

f (x)

x x

f (x)

f (x)

Suppose

(5)

(x)

x 1 x

x

f (x)

Then we have from Theorem 2.14

(6)

dt

a f (t)

x 1

f (x)

Hence from (5) and (6)

(x) f t dt.

a ( )

Also we have from Theorem 2.14

x

f (t)dt xf (x)

a

n n

And since f (x) is increasing, we get f (k) f (x)dx h(n) nf (n)

k 1 a

(7)

Given that

f (an )

f (n)

n n

n n

f (ak ) f (k)

By Result 3.5

n

n

f (ak ) nf (n)

By (7)

k 1

f (a ) nf (a )

k 1

f (a ) (a ) f (a )

k 1

lim

(an ) 1

k n

ak an

ak an

k n n

n

f (ak )

(8)

If

an x an1 we have

(an ) x an1

ak an

f (an )

And

f (an ) f x f an1

f (ak ) f (ak )

f (ak )

ak an ak x ak an1

lim

(an )

lim x lim

an

(a

a )

n

f (a )

x

f (a )

n

f (a )

n1 n

k k k

k k k

ak an ak x ak an

f (an )

x

f x

f an

x

f x

f x

f (ak )

1 lim

x

1

f (ak )

By (8)

lim

x

1

f (ak )

x ak x

ak x

f x

f (ak )

ak x

f x

x

f (x)

a x

k

k

f x

ak x

f (ak ) x

7. Theorem. Let f F,

an satisfies (1) and

f (an ) lf (n). Then

1 1 1

l

l

x

11

1 1 1

n

n

a ns f (n) (x)

l s xs

1

s

(x)

s

t

t

s

s

1 dt f (ak )s l s xs .

f (x)s

a f (t)s

ak x

Proof: Given an satisfies (1) and f F

(9)

and

f (an ) lf (n)

Suppose

(x)

1 1

l s xs

1

f (x)s

(an )

1 1

l s an s

1

f (an )s

1 1

l s a s

n n

n n

1

f (an )s

(9)

ns

lan

f (an )

an

• ns f (n) By

Conversely suppose

a ns f (n)

1 1

an s nf (n)s

lim

n

n

n

(an ) 1

1 1

l sans

By (9)

1

f (an )s

(10)

If an x an1 we have

(an ) x an1

We have by Theorem 2.6

increasing

f (x) x

has negative derivative

f (x) x

is decreasing

x is

f (x)

an

x

an1

lim (an )

lim x

lim an

(a a )

f (a ) f (x) f (a )

n 1 1

x 1 1

n 1 1

n1 n

n n1

l s a s

l s xs

l s a s

n n

1

f (an )s

1

f (x)s

1

f (an )s

1 lim x 1

x

lim 1

(x)

1 1

l s xs

x

1 1

l s xs

1

By (10)

x

1 1

l s xs

1

1

f (x)s

f (x)s f (x)s

s

s

an n f (n) (x)

1 1

l s xs

1

f (x)s

(11)

x

t f (t) dt

x x 1 f (x)

We have from Theorem 2.11

lim a 1

1

1

t f (t) dt

x x f (x)

1

a 1

x 1 1

1 1 1 1

In above equation put 1 1

and 1

1

t s f (t) s dt

s f (x) s

s s a

1 1 1

s

x

x

x 1 1 1

1 x 1 1 1 1

1 t s

s xs

1 dt 1

l s t s

s

1 dt

l s xs

1

a

(12)

f (t)s

f (x)s

1

l s x

1 1

t s

a f (t)s

f (x)s

From (11) and (12)

(x)

1 dt

s

s

a f (t)s

Also we have from Theorem 2.14

n

x

f (t)dt xf (x)

a

n

and

f (x) is increasing

Now

f (k) f (x)dx h(n) nf (n)

k 1 a

(13)

Given that

f (a)n lf (n)

n n

n n

f (ak ) l f (k)

By Result

k 1

3.5

k 1

n

n

f (ak ) lnf (n)

k 1

By (13)

ak an

f (ak ) lnf (an )

ak an

f (ak ) l (n) f (an )

f (ak )

(a )

(a ) ak an

n

n

lim n 1

(14)

lf (an )

n

f (ak )

If an x an1 we have

ak an

lf (an )

(an ) x an1

And

f (an ) f x f an1

lf (an ) lf x lf an1

f (ak ) f (ak )

f (ak )

ak an ak x ak an1

lim

(an )

lim x lim

an

(a

a )

n

f (a )

x

f (a )

n

f (a )

n1 n

k k k

k k k

ak an ak x ak an

lf (an )

lf x

lf an

1 lim

x 1

By (14)

lim

x 1

x

f (ak )

x

f (ak )

ak x

lf x

x

x

f (ak )

ak x

lf x

k

k

f (ak )

ak x

lf x

1 1 1

x a x

f (a ) lx

f (a )s

l s xs .

f (x) lf x

k

ak x

k

ak x

8. Theorem. If

g(x) is a f.s.i. and

g(a ) lg(n) where a

satisfies (1), then

n n

k

k

g(a )

(x) ak x , for all real .

g(x)

Proof: Given that g(x) is a f.s.i. and

g(a ) lg(n) where a

satisfies (1)

We have from Theorem 2.14

n

n

n

n

x

g(t) dt xg(x)

a

n n

And since g(x) is increasing, we get

g(k) g(x) dx h(n) ng(n)

(15)

Given that (16)

k 1

g(an ) lg(n)

a

n

n

g(a ) l g(n)

g(a ) l g(n)

g(a ) l g(n)

n n

k

By Result 3.5

k 1 k 1

g(a ) l ng(n)

g(a ) l ng(n)

n

k

k 1

By (15)

g(a ) ng(a )

k n

k n

ak an

g(a )

By (16)

g(a ) (a )g(a )

k

n

n

(a ) ak an

lim (an ) 1

ak an

k n n

n g(a )

n

g(a )

k

k

ak an

n

n

g(a )

(17)

If an x an1 and 0 (an ) (x) (an1 )

We have g F

g(a ) g(x) g(a ) g(a ) g(x) g(a )

n n1 n n1

g(a )

g(a )

g(a )

ak an

k k k ak x ak an1

k k k

k k k

g(a )

g(a )

g(a )

ak an

ak x

ak an1

n

n

g(an1 )

g(x)

g(a )

lim

(an )

lim (x) lim

(an )

(a

a )

n

g(a )

x g(a )

n

g(a )

n1 n

k k k

ak an ak x ak an

g(a ) g(x) g(a )

n n

1 lim (x) 1

x g(a )

k

k

By (17)

lim (x) 1

x g(a )

k

k

ak x

g(x)

ak x

g(x)

g(a )

k

k

(x)

ak x

g(x) .

If an x an1 and 0

(an ) (x) (an1 )

We have g F

g a g( x ) g a( )

g a ( g x) g a( )

( 0)

n n1

n1 n

lim

(an )

lim (x) lim

(an )

(a a )

n

g(a )

x g(a )

n

g(a )

n n1

k k k

ak an ak x ak an

g(a ) g(x) g(a )

n n

1 lim (x) 1

x g(a )

k

k

By (17)

lim (x) 1

x g(a )

k

k

ak x

g(x)

ak x

g(x)

k

k

g(a )

(x) ak x

Hence

g(x) .

k

k

g(a )

(x) ak x

g(x)

For all

real.

9. Theorem. If an satisfies (1) and (x) 1, the number of primes up to x, then

pn x

(x) (x).

a

a

k

k

Proof: The ak x are a1, a2,…, a ( x) .

Let us write

k x, (k 1, 2,…, (x))

log ak

log x

k log ak

log x

k

k

k

log x

log ak

(k 1, 2,…, (x))

We know that

(x) (x)

( x)

k

( x)

k

( x) 1

log x log a

k 1

(18)

k 1

1 1

k 1 k

From theorem 3.2, We have logan s log n

log an s log n

( x) 1 1

1 ( x) 1

log a

• log a s log k

By Result 3.5 (19)

k 1 k

1 k 2

x 1 x

2

2

x 1 x

We have from Theorem 2.14

(20)

dt

a f (t)

f (x)

log tdt log x

1 1 ( x) 1

( x) 1

(x)

Now

(20)

log a1

s k 2 log k

dt (1) By

2 log t s log x

( x) 1

(x)

From Equation (19) and above equation

log a

• s log x

k 1

k

h(x) (x) log x

From Equation (18) and above equation

(x) (x)

s log x

(h(x) 1)

1 (x) h(x) log x 1 lim (x) lim h(x) log x

(x)

(21)

s log x

x (x)

x

s log x

We have from Theorem 3.4 of (i)

log (x) 1 log x

lim

log x 1

s

(x)

x s log (x)

(x)

x

x

x

x

Using this inEquation (21), we get

1 lim (x) 1

(h(x) 1)

lim (x) 1

Hence

10. Theorem. Let f F, an satisfies (1).

(x) (x).

n ns 1 f (n) n

Then (i)

ak

k 1

s 1

• s 1 an

( 0)

(ii)

a (x) x

( 0)

n

an x

s 1

Proof: Given that Let f F, an satisfies (1)

n

n

1. Let us consider the sum 1 2 … (n 1) (ks f (k)s )

k n

(22)

n

n

n

n

Where n is positive integer in the interval a,

From Equation (1) we have (23)

a ns f (n)

a ns f (n) ( 0)

n n

We know that xs f (x)s is increasing

(ks f (k)s ) xs f (x) dx (ns f (n)

(24)

x

k n n

t f (t) dt

x x 1 f (x)

From Theorem 2.13, we have

lim

x

a

x 1 f (x)

1

t f (t)

a

dt

1

1

Put s

(25)

and

in above equation, we get

n

xs f (x) dx

n

ns 1 f (n)

s 1

From Eqations (22), (24) and (25), we get

n

n

1 2 … (n 1)

k n

s 1

n f (n)

n f (n)

(ks f (k)s )

s 1

n nf (n)

s

s

s 1

) (26)

s s

na

n

n

n

n

1 2 … (n

1) (k f (k) )

k n

s 1

By (23) (27)

FromEquations (23), (26) and (27) and using Result 3.5, we get

n ns 1 f (n) n

ak

k 1

s 1

• s 1 an

( 0)

1. If

an x an1 and 0

(an ) x an1

And

a x a

(s +1) a (s +1) a (s +1) a

n n1

k k k ak an ak x ak an1

lim

(an )

lim (x) lim

(an )

( a a )

n (s +1)

a

x (s +1) a

n

(s +1) a

n n1

k k k

ak an ak x ak an

a x a

n n

(28)

k

k

n na

(n)a

s 1 a

a

a

We have

a n

a n

(a )

ak an

k

k 1

s 1

k

ak an

s 1 n

(x)

lim

n

(an )

k

k

n

n

s 1 a =1

ak an

a

a

n

x

x

(x)

Equation (28) implies

1 lim (s +1) a 1

k

k

lim (s +1) a 1

k

k

x

x

ak x

x

ak x

x

k

k

(s +1) a

(x)

(x) ak x

x

a

n

n

an x

s 1

x . ( 0).

We apply the results discussed in this article to look into some of the applications in number theory.

Acknowledgments. The author would like to thank the anonymous referees for a careful reading and evaluating the original version of this manuscript.

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1

2

3

Appl.

4

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fourth Edition, 1960.

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R. Jakimczuk. A note on sums of powers which have a fixed number of prime factors, J.Inequal. Pure

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R. Jakimczuk. The ratio between the average factor in a product and the last factor, mathematical

sciences:

Quarterly Journal 1 (2007), 53-62.

5 Y. Shang, On a limit for the product of powers of primes, Sci. Magna 7 (2011) 31-33.

6 J. Rey Pastor, P. Pi Calleja and C. Trejo, An alisis Matem arico, Volumen I, Octava Edicion, Editorial Kapelusz,

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