VOL. 8, NO. 7, JULY 2013
ISSN 1819-6608
ARPN Journal of Engineering and Applied Sciences
©2006-2013 Asian Research Publishing Network (ARPN). All rights reserved.
www.arpnjournals.com
SLOW INCREASING FUNCTIONS AND THEIR APPLICATIONS TO
SOME PROBLEMS IN NUMBER THEORY
K. Santosh Reddy, V. Kavitha and V. Lakshmi Narayana
Vardhaman College of Engineering, Shamshabad, Hyderabad, Andhra Pradesh, India
E-Mail: santureddyk@gmail.com
ABSTRACT
This article commences with a definition of slow increasing function and moves on to delineate a few properties
of slow increasing functions. Besides, several applications in some problems of number theory using the theory of slow
increasing functions are also presented to show how useful these functions prove in solving complex problems.
Keywords: slow increasing functions, asymptotically equivalent, sequence of positive integers.
1. INTRODUCTION
Slow increasing functions are defined as follows:
1.1. 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
Note
Write
F = { f : f is a s.i.f.}.
lim
x →∞
xf ′( x)
=0
f ( x)
Note
that
lim
x→∞
xh′( x)
xα f ( x)α −1 f ′( x) = α lim xf ′( x) = 0.
= xlim
x →∞ f ( x)
→∞
h( x)
f ( x)α
(vi) Let h = f ο g i.e h( x) = f ( g ( x))
lim f ( x ) = xlim
log x = ∞
→∞
and
lim
α > 0, c > 0 be two
f + c (ii) f − c (iii) cf (iv) fg (v) f
(vii) log f (viii) f + g all lie in F .
(i)
Given
constants.
that
f , g ∈ F and
α
(vi)
fοg
α > 0, c > 0
Proof of (i), (ii), (iii), and (iv) follows the definition 1.1
(v) Let h = f α
h′ is continuous
xh′( x)
xf ′( g ( x)) g ′( x)
g ( x) f ′( g ( x)) xg ′( x)
= lim
= xlim
×
= 0.
→∞
h( x) x→∞
f ( g ( x))
f ( g ( x))
g ( x)
(vii) Let
h = log f
Note
that
h′( x) =
f ′( x)
> 0, and h′ is continuous
f ( x)
be
lim h( x) = xlim
log f ( x) = ∞,
→∞
x →∞
and
Moreover
f ′( x)
x
xh′( x)
f ( x)
f ′( x)
1
lim
x
= xlim
= xlim
×
= 0.
x →∞ h( x)
→∞ log f ( x)
→∞
f ( x) log f ( x)
Hence
Proof
and
and
Hence h = f ο g ∈ F
2. SOME PROPERTIES
and let
lim h( x) = xlim
f ( g ( x)) = ∞,
→∞
x →∞
Moreover
x →∞
f ( x) = log log x, x > e is also a s.i.f.
2.1. Theorem
Let f , g ∈ F
constants then we have
that
h′( x) = f ′( g ( x)) g ′( x) > 0,
1
f ′( x) = , ∀x > 1 and f ′ is continuous
x
xf ′( x )
1
x
Moreover lim
= lim ×
=0
x →∞ f ( x )
x →∞ x
log x
(ii)
f ′( x) > 0, and h′ is continuous
Hence h = f α ∈ F
x > 1 is a s.i.f.
x →∞
and
Moreover
Note
1.2. Examples
(i) f ( x) = log x,
lim h( x) = xlim
f ( x)α = ∞,
→∞
x →∞
α −1
h′( x) = α f ( x)
x →∞
increasing function (s.i.f. in short) if
that
h = log f ∈ F
(viii) Let
h= f +g
For sufficiently large x , we have 0 ≤
0≤
xf ′
xf ′
and
≤
f +g
f
xg ′
xg ′
≤
f +g
g
480
VOL. 8, NO. 7, JULY 2013
ISSN 1819-6608
ARPN Journal of Engineering and Applied Sciences
©2006-2013 Asian Research Publishing Network (ARPN). All rights reserved.
www.arpnjournals.com
xh′(x)
xf ′
xg′
≤ xlim
+ xlim
=0
By adding the above, we get 0 ≤ xlim
→∞ h( x)
→∞ f
→∞ g
Let h( x) = f ( x) and
g ( x)
xh′( x)
∴ xlim
= 0 Hence h = f + g ∈ F
→∞ h( x)
f , g ∈ F . Define h( x) = f ( xα ) and
k ( x) = f ( xα g ( x)) for each x, then h, k ∈ F .
⎛ f ′( x) g ( x) − f ( x) g ′( x) ⎞
x⎜
⎟
g ( x) 2
xh′( x)
⎠ = lim f ′( x) − lim xg ′( x) = 0
= lim ⎝
lim
x →∞ h( x )
x →∞
x →∞ f ( x )
x →∞ g ( x )
⎛ f ( x) ⎞
⎜ g ( x) ⎟
⎝
⎠
Hence
Proof
f , g ∈ F . Define h( x) = f ( xα )
and k ( x) = f ( xα g ( x)) for each x.
Given that
h( x) =
Note
f ( xα )
that
f ′( x) g ( x) − f ( x) g ′( x)
g ( x)2
Moreover
2.2. Theorem
Let
Let
h′( x) =
f
g
∈F
2.4. Theorem
Let
h : ⎡⎣a, ∞ ) → ( 0, ∞ )
be a continuously
differentiable function such that h′( x ) > 0 and lim h( x) = ∞
x →∞
lim h( x) = xlim
f ( xα ) = ∞,
x →∞
→∞
and
h′( x) = f ′( xα )α xα −1 > 0, and h′ is continuous
(i) Define g ( x) = h(log x) . Then g ∈ F ⇔ lim h′( x) = 0
x→∞ h( x)
(ii) Define k ( x) = e h ( x ) . Then
k ∈ F ⇔ lim xh′( x) = 0
x →∞
Moreover
lim
x →∞
xh′( x)
xf ′( xα )α xα −1
xα f ′( xα )
α
lim
= xlim
=
=0
→∞
x →∞
h( x )
f ( xα )
f ( xα )
Hence
Let
h( x) = f ( xα ) is s.i.f.
k ( x) = f ( xα g ( x))
Note that
lim k ( x) = lim f ( xα g ( x)) = ∞, and
x →∞
x →∞
k ′( x) = f ′( xα g ( x)) ⎡⎣α xα −1 g ( x) + xα g ′( x) ⎤⎦ > 0 and
k′ is continuous
Proof
Given that h′( x) > 0 and lim h( x) = ∞
x →∞
(i) Define g ( x) = h(log x) then
g∈F
Suppose
g ′( x) =
then g satisfies
h′(log x)
x
lim
x →∞
xg ′( x)
=0
g ( x)
h′(log x)
h′(log x)
x
=0
i.e. lim
= 0 ⇒ lim
x →∞ h(log x )
x →∞ h(log x)
x
h′(t )
=0
t →∞ h(t )
Put t = log x so that x → ∞ ⇒ t → ∞ ∴ lim
Moreover
xf ′( xα g ( x)) ⎡⎣α xα −1 g ( x) + xα g ′( x) ⎤⎦
xh′( x)
= lim
x →∞ h( x )
x →∞
f ( xα g ( x))
lim
xα g ( x ) f ′( xα g ( x ))
xα g ( x) f ′( xα g ( x)) xg ′( x)
= α lim
+ lim
×
=0
α
x →∞
x →∞
f ( x g ( x ))
f ( xα g ( x))
g ( x)
α
Therefore k ( x) = f ( x g ( x)) is s.i.f. Hence h, k ∈ F
2.3. Theorem
Let
f ,g∈F
be
such
that
f
f ( x)
d ⎡ f ( x) ⎤
= ∞ and
> 0. Then ∈ F .
lim
⎢
⎥
x →∞ g ( x )
g
dx ⎣ g ( x) ⎦
h′( x)
=0.
x →∞ h( x )
i.e. lim
h′( x)
=0
h( x )
x
Put t = e so that x = log t and x → ∞ ⇒ t → ∞
h′( x)
h′(log t )
⇒ lim
= lim
=0
x →∞ h( x )
t →∞ h(log t )
Conversely suppose lim
x →∞
tg ′(t )
h′(log t )
h′( x)
= lim
= lim
= 0.
t →∞ g (t )
t →∞ h(log t )
x →∞ h( x )
Now lim
Hence g ∈ F
(ii) Like proof of (i)
Proof
Given that
f ( x)
d ⎡ f ( x) ⎤
f , g ∈ F , lim
= ∞ and
>0
x →∞ g ( x )
dx ⎢⎣ g ( x) ⎥⎦
2.5. Theorem
log f ( x)
= 0.
x →∞
log x
If f ∈ F then lim
481
VOL. 8, NO. 7, JULY 2013
ISSN 1819-6608
ARPN Journal of Engineering and Applied Sciences
©2006-2013 Asian Research Publishing Network (ARPN). All rights reserved.
www.arpnjournals.com
For any α with 0 < α <
Proof
Given
that
f ∈ F,
⎛ f ′( x) ⎞
⎜ f ( x) ⎟
log f ( x)
⎠ (byL’Hospital’s rule)
lim
= lim ⎝
x →∞
x →∞
log x
⎛1⎞
⎜ ⎟
⎝ x⎠
xf '( x)
log f ( x)
= lim
= 0 i.e. lim
= 0.
x →∞ f ( x )
x →∞
log x
2.6. Theorem
f ∈ F if and only if to each α > 0 there
d ⎡ f ( x) ⎤
exists xα such that
< 0, ∀x > xα
dx ⎢⎣ xα ⎥⎦
Proof
We have
⎤
d ⎡ f ( x) ⎤ f ′( x) xα − f ( x)α xα −1 f ( x) ⎡ xf ′( x)
=
= α +1 ⎢
−α ⎥
α ⎥
2α
⎢
dx ⎣ x ⎦
x
x
⎣ f ( x)
⎦
Suppose f ∈ F
xf '( x)
=0
x →∞ f ( x )
then ⇒ lim
i.e. For each α > 0 there exists xα such that ∀x > xα
d ⎡ f (x) ⎤
xf '(x)
And
− 0 < α, ∀x > xα ⇒ ⎢ α ⎥ < 0, ∀x > xα
dx ⎣ x ⎦
f (x)
To prove the converse assumes that the condition
holds.
Let α
> 0 be given. Then there exist xα such
d ⎡ f ( x) ⎤
< 0, for all ∀x > xα for some xα
dx ⎢⎣ xα ⎥⎦
f ( x)
is decreasing for x > xα
This implies that
xα
f ( x)
bounded above, say, by M
Hence
xα
That is, there exists M > 0 such that
f ( x)
0 < α < M , ∀x > xα
x
f ( x)
f ( x) 1
lim β = lim α β −α = 0
x →∞ x
x →∞ x
x
2.8. Note
We know that each f ∈ F is an increasing
function. Moreover, by the above theorem, it is clear that
f ( x)
= 0, ∀β >0 . This shows that the increasing
xβ
nature of f is slow. In other words, f does not increase
lim
x →∞
rapidly. This justifies the name given to the members of F.
From the above theorem, we have the following
results:
2.9. Theorem
If
d ⎡ f ( x) ⎤
< 0 this implies that
dx ⎢⎣ xα ⎥⎦
lim
x →∞
f ( x)
=0
x
Proof
x →∞
xf ′( x)
=0
x →∞ f ( x )
f ( x)
= 0.
x
If f ∈ F , then lim
x →∞
xf ′( x)
xf '( x)
= 0.
→ 0 as x → ∞ ⇒ lim
x →∞ f ( x )
f ( x)
Therefore f ∈ F .
and
x →∞
Since lim
i.e.
f ( x)
=0 we must have lim f ′( x)=0.
x →∞
x
2.10. Theorem
Let f ∈ F , then for any α > −1and β ∈ , the
∞
series
∑ nα f (n)β
diverges to ∞ .
n =1
2.7. Theorem
Proof
then
In Theorem 2.7 put β = 1 , toget lim
xf '( x)
− 0 < α , ∀x > xα
f ( x)
If f ∈ F then
f ∈F
lim f ′( x) = 0.
that ∀x > xα
We have, by hypothesis
β , we get by Theorem 2.6,
lim
x→∞
f (x)
= 0 , for all β > 0
xβ
Proof
∞
We write
∞
∑ nα f (n) β = ∑ ( nα +1 f (n) β )
n =1
n =1
∞
1
we know that the series ∑ diverges to ∞
n =1 n
482
1
n
VOL. 8, NO. 7, JULY 2013
ISSN 1819-6608
ARPN Journal of Engineering and Applied Sciences
©2006-2013 Asian Research Publishing Network (ARPN). All rights reserved.
www.arpnjournals.com
Given α > −1 ⇒ α + 1 > 0
x →∞
If β ≥ 0 then lim nα +1 f (n) β = ∞
n →∞
If β > 0 then
n
nα +1
= lim
= ∞ (from Theorem 2.7)
−
β
n →∞ ⎛ f ( n)
⎞ n→∞ f (n) − β
⎜
⎟
α
⎝ n
⎠
lim
∞
∑ nα f (n)β
i.e.
f ( x)
= 1, then f is said to asymptotically
g ( x)
equivalent to g . We describe this by writing f g .
(ii) f = Ο ( g ) Means f ≤ Ag for some A > 0. In this
case we say that f is of large order g .
f ( x)
(ii) f = ο ( g ) Means lim
= 0 . In this case we
x →∞ g ( x )
say that f is of small order g .
(i) If lim
diverges to ∞
n =1
2.13. Examples
An important byproduct of the above theorem is
the following result.
(i) Consider f ( x ) = x , g ( x) = x + x, for all
2.11. Theorem
Let f ∈ F . Then for any
f ( x)
x
= lim n
=1
→∞
x
g ( x)
x +x
Therefore f g .
n
n
x>0
n
x
lim
x →∞
∫t
α
α > −1
and β ∈ ,
f (t ) β dt
= 1.
a
α +1
β
⎛ x f ( x) ⎞
⎜
⎟
⎝ α +1 ⎠
lim
x →∞
(ii) x = Ο (10 x ) Because
(iii) x + 1 = ο ( x
Proof
From
Theorem
2.10,
we
n →∞
xα +1
f ( x) β = ∞, ∀ α > −1, ∀β
x →∞ α + 1
⇒ lim
∞
Theorem
2.10,
we
have
2
∑ tα f (t )β = ∞
x
1
1
= ⇒ x = (10 x).
10 x 10
10
x +1
= 0.
x →∞ x 2
) Because lim
have
lim xα +1 f ( x) β = ∞
From
and
As a result of Theorem 2.11, we get the following
results as particular cases.
2.14. Theorem
Let f ∈ F . Then we have the following
statements.
t =1
x
x
(i)
⇒ lim ∫ t f (t ) dt = ∞
x →∞
α
β
Consider lim
x →∞
∫t
α
f (t ) β dt
a
β
x f (x)
x→∞
xα+1
xα f (x)β +
β f (x)β −1 f ′(x)
α +1
= lim
α
∫
x
f (t ) dt xf ( x) (iii)
a
1
∫ f (t ) dt a
x
f ( x)
Proof
(By L’Hospitals’s rule)
β
x f ( x)
=1
x →∞
β xf ′( x) ⎞
α
β ⎛
x f ( x) ⎜ 1 +
⎟
⎝ α + 1 f ( x) ⎠
= lim
dt xf ( x) β
x
(ii)
⎛ xα +1 f ( x ) β ⎞
⎜
⎟
⎝ α +1 ⎠
α
β
a
a
x
∫ f (t )
Let f ∈ F
(i) Put α = 0 in Theorem 2.11, we get
x
lim
x →∞
∫ f (t )
a
β
dt
xf ( x) β
x
= 1 ⇒ ∫ f (t ) β dt xf ( x) β
a
(ii) Put α = 0, β = 1 in Theorem 2.11, we get
2.12. Definition
Let
f , g : [ a, ∞ ) → ( 0, ∞ )
483
VOL. 8, NO. 7, JULY 2013
ISSN 1819-6608
ARPN Journal of Engineering and Applied Sciences
©2006-2013 Asian Research Publishing Network (ARPN). All rights reserved.
www.arpnjournals.com
x
∫ f (t )dt
lim a
x →∞
xf ( x)
x
= 1 ⇒ ∫ f (t )dt xf ( x)
a
(iii) Put α = 0, β = -1 in Theorem 2.11, we get
x
lim
1
∫ f (t )dt
a
x
f ( x)
x →∞
x
=1
⇒∫
a
1
x
dt f (t )
f ( x)
lim
x →∞
f ( x + c)
= 1, For any c ∈ f ( x)
lim
x →∞
f (cx)
= 1, for any
f ( x)
Let f ∈ F
f ( x + c ) − f ( x) = ( x + c − x) f ′(t )
f ( x + c ) − f ( x) cf ′(t )
0≤
=
f ( x)
f ( x)
f ( x + c) − f (x)
cf ′(t) , t ∈ x, x + c
(
)
= lim
x→∞ f ( x)
f (x)
f ( x +c)
⇒ lim
−1= 0, since limf ′(x) =0 (by Theorem 2.9)
x→∞
x→∞ f (x)
0 ≤ lim
x→∞
lim
x →∞
f ( x + c)
= 1.
f ( x)
0 ≤ lim
x→∞
f ( x) − f ( x + c )
cf ′(t )
=−
f ( x)
f ( x)
f (x) − f ( x + c)
f (x)
f ( cx ) − f ( x )
f ( x)
= (c − 1) lim
x →∞
And f ( x ) is decreasing ⇒
xf ′(t )
f ( x)
,
t ∈ ( x, cx )
f ′( x) > f ′(t )
f ( cx)
−1=0, since lim f ′(x) = 0 (by Theorem 2.9)
There fore lim
x→∞ f (x)
x→∞
lim
x →∞
f ( cx )
= 1.
f ( x)
Case (b). Suppose c < 1
By Lagrange’s mean value theorem there
exists t ∈ ( cx, x ) such that
f ( x) − f ( cx ) = ( x − cx) f ′(t )
⇒
0≤
⇒ 0 ≤ lim
f ( x) − f ( cx ) (1 − c) xf ′(t )
=
f ( x)
f ( x)
f ( x ) − f ( cx )
f ( x)
x →∞
= (1 − c ) lim
x →∞
And f ( x ) is decreasing ⇒
xf ′(t )
f ( x)
, t ∈ ( cx, x )
f ′( x ) > f ′(t )
f ( cx )
−1 = 0 ,
f ( x)
(by Theorem 2.9)
since lim f ′( x ) = 0
x →∞
⇒ lim
x →∞
f ( x) − f ( x + c ) = ( x − x − c) f ′(t )
⇒
f ( cx ) − f ( x) (c − 1) xf ′(t )
=
f ( x)
f ( x)
0≤
x →∞
exists t ∈ ( x + c, x ) such that
0≤
exists t ∈ ( x, cx ) such that
There fore lim
Case (b). Suppose c < 0
By Lagrange’s mean value theorem there
⇒
(ii) Case (a). Suppose c > 1
By Lagrange’s mean value theorem there
⇒
a t ∈ ( x, x + c ) such that
⇒
x →∞
(by Theorem 2.9)
f ( x + c)
= 1.
f ( x)
x →∞
(i) Case (a). Suppose c > 0
By Lagrange’s mean value theorem, There exists
⇒
lim
⇒ 0 ≤ lim
c∈
⇒
⇒
⇒
(ii) If f ′( x ) is decreasing then
Proof
f ( x+c)
−1=0, since lim f ′(x) = 0
x→∞
f (x)
x→∞
f ( cx ) − f ( x) = (cx − x) f ′(t )
2.15. Theorem
Let f ∈ F . Then
(i)
⇒ lim
= −c lim
x→∞
f ′(t) , t ∈ x + c, x
(
)
f (x)
f ( cx )
= 1.
f ( x)
2.16. Theorem
Suppose f ∈ F
is
such
that
f ′( x )
decreasing. If 0 < c1 ≤ c2 and
g is a function such that
f ( g ( x ) x)
c1 ≤ g ( x) ≤ c2 then lim
= 1.
x →∞
f ( x)
484
is
VOL. 8, NO. 7, JULY 2013
ISSN 1819-6608
ARPN Journal of Engineering and Applied Sciences
©2006-2013 Asian Research Publishing Network (ARPN). All rights reserved.
www.arpnjournals.com
Proof
If 0 < c1 ≤ g ( x) ≤ c2 ⇒ f (c1x) ≤ f (g(x)x) ≤ f (c2 x)
since f is decreasing
f (c1 x) f ( g ( x) x) f (c2 x)
≤
≤
f ( x)
f ( x)
f ( x)
f (c1 x)
f (c2 x)
f ( g ( x) x)
lim
≤ lim
≤ lim
x →∞ f ( x )
x →∞
x →∞ f ( x )
f ( x)
f ( g ( x) x)
1 ≤ lim
≤1
(By Theorem 2.15)
x →∞
f ( x)
f ( g ( x) x)
lim
= 1.
x →∞
f ( x)
⇒
⇒
⇒
⇒
3. APPLICATIONS OF SLOW INCREASING
FUNCTIONS TO SOME PROBLEMS OF
NUMBER THEORY
This section details some applications in
problems pertaining to number theory.
We begin with the following important definition.
3.1. Definition
Let f ∈ F . Through out
( an ) denotes a strictly
increasing sequence of positive integers such that
a1 > 1 and lim
n →∞
an
= 1 for some s ≥ 1.
n f ( n)
And
an n s f (n)
By prime number theorem we have lim
n →∞
=1.
pn
=1
n log n
n′ is positive integer in interval [ a, ∞ )
If
Then
n
∑ log a
k
k = n′
f : ( a, ∞ ) → (1, ∞ ) be a s.i.f. (a > 1)
tf ′(t ) β
dt = ∞ ( a < b ) . Suppose (an ) be
x →∞ ∫ f (t )
b
an n f (n)
s
of
positive
( s ≥ 1) .
Then
a .a ...a
1
lim 1 2 n = s .
n →∞
an
e
n
Proof
integers
n
n
k = n′
k = n′
k = n′
Since log x is increasing and positive in
(2)
( a, ∞ )
Now
n
n
k = n′
n′
∑ log k = ∫ log xdx + Ο(log n)
= n log n − n + Ο(log n)
= n log n − n + ο (n)
(3)
On the other hand if
ε >0
we have for all
k =n′
n
n
≤
∑ ο (1)
k =n′
such
n
<
ε (n − n′ + 1)
n
≤ 2ε
n
i.e.
∑ ο (1) = ο (n)
(4)
k = n′
n
n
k = n′
n′
∑ log f (k ) = ∫ log f ( x)dx + Ο(log f (n))
x
sequence
n
= s ∑ log k + ∑ log f (k ) + ∑ ο (1)
We find that
and lim
the
( s ≥ 1)
⇒ logan = s log n + log f (n) + ο (1)
∑ο (1)
There exist several such sequences.
For example an = pn , the sequence of prime
Let
(a > 1) and lim ∫
n
i.e. an n s f (n)
3.2. Theorem
tf ′(t ) β
dt = ∞ ( a < b )
x →∞
f (t )
b
x
n ≥ n′ the inequality ο (1) < ε
Therefore for n > n′ , we have
s
numbers in increasing order, f ( x ) = log x and s
f : ( a, ∞ ) → (1, ∞ ) be a s.i.f.
Given that
Suppose f ∈ F is such that f ′( x) is decreasing
xf ′( x)
dx + Ο(log f ( n))
f ( x)
n′
n
= n log f ( n) − ∫
that
(1)
(5)
We know that
log f (n)
f ′(n)
= lim
n →∞
x →∞ f ( n)
n
lim
= lim
n →∞
(By L’Hospital’s rule)
nf ′(n) 1
× = 0.
f ( n) n
485
VOL. 8, NO. 7, JULY 2013
ISSN 1819-6608
ARPN Journal of Engineering and Applied Sciences
©2006-2013 Asian Research Publishing Network (ARPN). All rights reserved.
www.arpnjournals.com
Ο(log f ( n)) = ο ( n)
⇒
(6)
Therefore
n
Now
tf ′(t )
∫n′ f (t ) dt
xf ′( x)
= lim
=0
lim
→∞
x →∞
x
x
f ( x)
x
⇒
n
n →∞
(By L’Hospital’s rule)
an
⇒
es
n
a1a2 ...an
1
s
an
e
a1a2 ...an 1
= s.
an
e
In view of the above theorem and prime number
theorem implies the following.
xf ′( x)
dx = ο (n)
f ( x)
n′
n
⇒
lim
a1a2 ...an ∫
(7)
3.3. Theorem
Let pn be the sequence of prime numbers. Then
From (5), (6) and (7), we get
n
∑ log f (k ) = n log f (n) + ο (n)
(8)
⇒
lim
n
n →∞
k = n′
Proof
From (2), (3), (4) and (7), we get
n
∑ log a
k =1
⇒
= s( n log n − n + ο (n)) + n log f ( n) + ο ( n) + ο (n)
k
k
k =1
= sn log n − sn + n log f (n) + ο (n) (9)
∑ log a
k =1
k
= log a1 + log a2 + ... + log an
⎛ n
⎞
= log a1a2 ...an ⇒ a1a2 ...an = exp ⎜ ∑ log ak ⎟
⎝ k =1
⎠
⇒
n
⎛
⎜
a 1 a 2 ...a n = ex p ⎜
⎜
⎜
⎝
n
∑ lo g a
k =1
n
k
⎞
⎟
⎟
⎟
⎟
⎠
⎛ n
⎞
⎜ ∑ log ak ⎟
sn log n − sn + n log f (n) + ο (n) ⎞ (By 9)
k =1
⎟ = exp ⎜⎛
⇒ exp ⎜
⎟
n
n
⎜
⎟
⎝
⎠
⎜
⎟
⎝
⎠
= exp ( log n s − s + log f (n) + ο (1) )
= exp ( log n s f (n) + ο (1) ) e − s
=
exp ( log n s f (n) + ο (1) )
es
n s f (n) an
s
es
e
f ( x ) = log x
and s = 1 .
Let cn , k be the sequence of integers which have
Jakimczuk
cn ,k But
n
In theorem 3.2 put an = pn ,
in their prime factorization
n
∑ log a
p1 p2 ... pn 1
= .
pn
e
[ 4] proved that
k prime factors. Rafael
( k − 1)! n log n
( log log n )
( k −1)
As a result of previous theorem, we have the
following result.
3.4. Theorem
lim
n
c1,k .c2,k ...cn , k
n →∞
cn , k
1
= .
e
Proof
In
Theorem
3.2
put
a n = cn , k ,
f ( x) = (k − 1)!log n and s = 1 .
CONCLUSIONS
We apply the results discussed in this article to
look into some of the applications in number theory.
ACKNOWLEDGEMENTS
The authors like to express their gratitude
towards the management of Vardhaman College of
Engineering for their continuous support and
encouragement during this work. Further authors would
like to thank the anonymous referees for going through the
article with a fine tooth comb and making critical
comments to the original version of this manuscript.
486
VOL. 8, NO. 7, JULY 2013
ISSN 1819-6608
ARPN Journal of Engineering and Applied Sciences
©2006-2013 Asian Research Publishing Network (ARPN). All rights reserved.
www.arpnjournals.com
REFERENCES
[1] G. H. Hardy and E. M. Wright. 1960. An Introduction
to the Theory of Numbers. Fourth Edition.
[2] R. Jakimczuk. 2010. Functions of slow increase and
integer sequences. Journal of Integer Sequences. 13,
Article 10.1.1.
[3] R. Jakimczuk. 2005. A note on sums of powers which
have a fixed number of prime factors. J. Inequal. Pure
Appl. Math. 6: 5-10.
[4] R. Jakimczuk. 2007. The ratio between the average
factor in a product and the last factor, mathematical
sciences: Quarterly Journal. 1: 53-62.
[5] Y. Shang. 2011. On a limit for the product of powers
of primes. Sci. Magna. 7: 31-33.
[6] J. Rey Pastor, P. Pi Calleja and C. Trejo. 1969. An
alisis Matem arico, Volume I, Octava Edition, Editorial
Kapelusz.
487