SOOCHOW JOURNAL OF MATHEMATICS
Volume 29, No. 4, pp. 353-361, October 2003
INEQUALITIES AND MONOTONICITY OF SEQUENCES
INVOLVING
n
(n + k)!/k!
BY
FENG QI
Abstract. Using Stirling’s formula, for all nonnegative integers k and natural
numbers n and m, it is proved that
n+k
i=k+1
1/n n+m+k 1/(n+m)
i
i
i=k+1
≤
n+k
.
n+m+k
From this, some monotonicity results of sequences involving
duced, and the related inequalities are refined.
n
(n + k)!/k! are de-
1. Introduction
In [7], H. Minc and L. Sathre proved that, if n is a positive integer, then
(n!)1/n
n−1
<
< 1.
n
[(n − 1)!]1/(n−1)
(1)
In [1, 6], H. Alzer and J. S. Martins refined the left inequality in (1) and showed
that, if n is a positive integer, then, for all positive real numbers r, we have
1/r
√
n
n
1
1 n+1
n!
n
r
r
.
(2)
<
i
i
< n+1
n+1
n i=1
n + 1 i=1
(n + 1)!
Received June 21, 2002; revised June 19, 2003.
AMS Subject Classification. Primary 05A10, 26D15; secondary 26A48, 33B15.
Key words. inequality, monotonicity, Stirling’s formula, sequence.
The author was supported in part by NNSF (#10001016) of China, SF for the Prominent
Youth of Henan Province (#0112000200), SF of Henan Innovation Talents at Universities, NSF
of Henan Province (#004051800), SF for Pure Research of Natural Science of the Education
Department of Henan Province (#1999110004), Doctor Fund of Jiaozuo Institute of Technology,
P.R. China.
353
354
FENG QI
Both bounds in (2) are the best possible.
Let n and m be natural numbers, k a nonnegative integer. The author
generalized in [11] the left inequality in (2) and obtained
n+k
<
n+m+k
1 n+k
ir
n i=k+1
1 n+m+k
ir
n + m i=k+1
1/r
,
(3)
where r is a given positive real number. The lower bound in (3) is the best
possible.
By mathematical induction, N. Elezović and J. Pečarić [3] and the author [9,
14] further generalized the left side of inequalities in (2) and the inequality (3) to a
large class of positive, increasing and convex sequences and of positive, increasing
and logarithmically concave sequences in different directions, respectively.
Recently, the inequalities in (1) were generalized by the author and others in
[5, 16, 17] and the following inequalities were obtained:
n
n+k+1
<
n+m+k+1
(n + k)!/k!
< 1,
(n + m + k)!/k!
n+m
(4)
where n and m are natural numbers, k is a nonnegative integer. Meanwhile, some
monotonicity results of the sequences involving n (n + k)!/k! were presented.
Some new results related to the Alzer’s and Martins’ inequality can also be
found in [2, 4, 10, 12, 15].
In this article, using inequalities deduced from Stirling’s formula, we will give
generalizations and refinements of the right inequality in (4), that is
Theorem 1. Let k be a nonnegative integer, n and m be natural numbers,
then
n+k 1/n n+m+k 1/(n+m)
n+k
.
(5)
i
i
≤
n+m+k
i=k+1
i=k+1
When n = m = 1, the equality in (5) is valid.
We also obtain related monotonicity results, as follows.
Theorem 2. The sequences
n+k
i=k+1
1/n i
√
n+k
(6)
INEQUALITIES AND MONOTONICITY OF SEQUENCES INVOLVING
n
(n + k)!/k!
355
are strictly increasing with n and k, respectively. The sequences
n+k
1/n n+m+k 1/(n+m)
i
i
i=k+1
(7)
i=k+1
are strictly increasing in k for all given natural numbers n and m.
It is well-known that factorials and their ‘continuous’ extension play an eminent role, for instance, in Combinatorics, Graph Theory, and Special Functions.
2. Lemmas
In order to verify our theorems, we need some lemmas.
Lemma 1. For all natural numbers n, we have
1
ln n − n + 1,
n ln n − n ≤ ln n! ≤ n +
2
(8)
n n
n n
√
√
1
1
< n! < 2πn
.
2πn
exp
exp
e
12n + 1
e
12n
(9)
For details about inequalities (8) and (9), please refer to [8, p.184 and p.194].
Lemma 2. For all natural numbers k and n > 1, we have
1
n(n − 1)
ln 1 +
2
n+k
+
2k + 1
n
ln 1 +
2
k
≤n+
12n − 1
. (10)
12(n + k)(12k + 1)
Proof. For x ≥ 2 and y ≥ 1, let
ϕ(x, y) =
1
2y + 1
x
x(x − 1)
ln 1 +
ln 1 +
+
2
x+y
2
y
12x − 1
− x.
−
12(x + y)(12y + 1)
(11)
Differentiating and simplifying produces
1
∂ϕ(x, y) 2x − 1
=
ln 1 +
∂x
2
x+y
6y 2 + 5y − 1 + (5 + 6y − 12y 2 )x − 30yx2 − 18x3
.
+
12(x + y)2 (x + y + 1)
(12)
356
FENG QI
Using the inequality
ln(1 + t) ≤
t(2 + t)
,
2(1 + t)
(13)
for t ≥ 0 in [13], we have
2y − 1 + 2x − 6yx2 − 6x3
∂ϕ(x, y)
≤
≤ 0,
∂x
12(x + y)2 (x + y + 1)
(14)
therefore ϕ(x, y) is decreasing with x ≥ 2.
It is easy to see that
1
2y + 1
2
ln 1 +
+
ϕ(2, y) = ln 1 +
y+2
2
y
1
23
ϕ(2, 1) = 2 ln 2 + ln 3 −
− 2 < 0.
2
468
−
23
− 2,
12(y + 2)(12y + 1)
Denote
ψ(y) = ϕ(2, y) +
then
23
,
12(y + 2)(12y + 1)
2
2y 2 + 8y + 3
,
−
ψ (y) = ln 1 +
y
y(y + 2)(y + 3)
18 − 6y − 5y 2
.
ψ (y) =
[y(y + 2)(y + 3)]2
(15)
(16)
(17)
Thus ψ (y) is decreasing with y ≥ 2. Since limy→∞ ψ (y) = 0, then ψ (y) ≥ 0 for
y ≥ 2 and ψ(y) increases for y ≥ 2.
Also since limy→∞ ψ(y) = 0, we get ψ(y) < 0 for y ≥ 2 and ϕ(2, y) < 0
for y ≥ 2. Hence the sequence {ϕ(2, k)}+∞
k=1 is negative, and so is the sequence
{ϕ(n, k)}+∞
n=2,k=1 . Lemma 2 follows.
3. Proofs of Theorems
Proposition 1. Let n be a natural number, then
√
n
n!
n
≤
.
n+1
n+1
(n + 1)!
When n = 1, the equality in (18) holds.
(18)
INEQUALITIES AND MONOTONICITY OF SEQUENCES INVOLVING
n
(n + k)!/k!
357
Proof. Let
f (x) = x(x − 1) ln(1 + x) − (x2 − x − 1) ln x − 2x + 2,
x ≥ 1.
(19)
By standard arguments, we get
1
x(x − 1) x2 − x − 1
−
−2
+
x
1+x
x
2x − 1 x(x − 1) x2 − x − 1
+
−
−2
≤
x
1+x
x
1−x
=
1+x
≤ 0,
f (x) = (2x − 1) ln 1 +
(20)
f (1) = 0.
Therefore, f (x) is decreasing and f (x) ≤ 0 for x ≥ 1. And for n ≥ 1, we obtain
n(n − 1) ln(1 + n) + 2 ≤ (n2 − n − 1) ln n + 2n,
(21)
this is equivalent to
n(n − 1)
1
n(n + 1)
ln n −
ln(1 + n).
n+
ln n − n + 1 ≤
2
2
2
(22)
Thus, from the right-hand side of inequality (8), we obtain
ln n! ≤
n(n − 1)
n(n + 1)
ln n −
ln(1 + n),
2
2
(23)
that is,
nn(n+1)/2
,
(n + 1)n(n−1)/2
(24)
nn(n+1)/2
n!
≤
,
(n + 1)n
(n + 1)n(n+1)/2
(25)
n! ≤
n
≥
n+1
(n!)n+1
[(n + 1)!]n
1/n(n+1)
.
(26)
The proof is complete.
Proposition 2. Let k be a nonnegative integer, and let n be a natural
number. Then
n+k
i=k+1
1/n n+k+1 1/(n+1)
i
i
i=k+1
≤
n+k
.
n+k+1
(27)
358
FENG QI
If n = 1, then equality in (27) holds for all such k.
Proof. The case k = 0 is Proposition 1, so we consider k ≥ 1.
For n > 1, the inequality (10) may be rewritten as
12n − 1
n(n − 1) n + k + 1 2k + 1 n + k
ln
+
ln
≤n+
,
2
n+k
2
k
12(n + k)(12k + 1)
n − n2 + 2k + 1
n(n − 1)
ln(n + k + 1) +
ln(n + k)
2
2
1
12n − 1
≤ k+
,
ln k + n +
2
12(n + k)(12k + 1)
1
1
1
1
−
n+k+
ln(n + k) − k +
ln k − n +
2
2
12(n + k) 12k + 1
n(n − 1)
n(n + 1)
ln(n + k) −
ln(n + k + 1).
≤
2
2
(28)
(29)
(30)
Then, substituting inequalities in (9) into the final inequality above yields
ln(n + k)! − ln k! ≤
n(n − 1)
n(n + 1)
ln(n + k) −
ln(n + k + 1),
2
2
(31)
this inequality can be rearranged as
(n + k)n(n+1)/2
(n + k)!
≤
,
k!
(n + k + 1)n(n−1)/2
n+k
i=k+1
i
(n + k + 1)n
≤
n+k
≥
n+k+1
n+k
n+k+1
n+k
(32)
n(n+1)/2
n n+k
(33)
n+k+1 n 1/n(n+1)
i
i=k+1
,
i
i
i=k+1
.
(34)
i=k+1
From this, the inequality (27) follows.
Proofs of Theorem 1 and Theorem 2. Proposition 2 implies
n+k
i=k+1
√
1/n
n+k+1 1/(n+1)
i
n+k
≤
i=k+1
√
i
n+k+1
,
(35)
n
INEQUALITIES AND MONOTONICITY OF SEQUENCES INVOLVING
1/n √
/ n
i=k+1 i)
n+k
and so (
(n + k)!/k!
359
+ k is increasing with n. Therefore
n+k
i=k+1
1/n
n+m+k 1/(n+m)
i
√
n+k
i=k+1
√
≤
i
.
n+m+k
(36)
Thus we have proved Theorem 1 and the first statement in Theorem 2.
It is easy to see that
n+k+1 1/n
i=k+2
i
√
=
n+k+1
n+k
n+k+1
n+k+1
k+1
1/n
n+k
1/n
i=k+1
√
i
.
(37)
x ≥ 1.
(38)
n+k
Let
g(x) =
1
1
1 1
−
ln(x + k + 1) − ln(k + 1) + ln(x + k),
x 2
x
2
Using
2t
≤ ln(1 + t),
2+t
(39)
for t > 0 in [8, p.273-274] and [13] and differentiating yields
k+1
1
x(3x + 2k)
+ ln
g (x) = 2
x 2(x + k)(x + k + 1)
x+k+1
2x
1
x(3x + 2k)
−
≤ 2
x 2(x + k)(x + k + 1) x + 2(k + 1)
2−x
.
=
2(x + k)(x + k + 1)[x + 2(k + 1)]
Thus, g(x) is decreasing with x ≥ 2.
1
2
ln
k+2
k+1
> 0, then the sequence
{g(n)}+∞
n=1
Since limx→∞ g(x) = 0 and g(1) =
is positive. Hence
n+k
n+k+1
n+k+1
k+1
n+k+1 1/n
i=k+2
√
i
(40)
1/n
n+k
i=k+1
> 1,
(41)
1/n
i
> √
.
n+k+1
n+k
1/n √
/ n + k being increasing with k are verified.
The sequences ( n+k
i=k+1 i)
(42)
360
FENG QI
It is clear that
n
(n + k + 1)!/(k + 1)!
(n + m + k + 1)!/(k + 1)!
n+m
(n + k + 1)n+m
=
(k + 1)m (n + m + k + 1)n
1/n(n+m)
n
(n + k)!/k!
.
(n + m + k)!/k!
n+m
(43)
Let
h(x) = (x + m) ln(x + k + 1) − x ln(x + m + k + 1) − m ln(k + 1),
Direct calculation leads to
m(2x + m + k + 1)
m
h (x) =
− ln 1 +
(x + k + 1)(x + m + k + 1)
x+k+1
m
m(2x + m + k + 1)
−
≥
(x + k + 1)(x + m + k + 1) x + k + 1
mx
=
(x + k + 1)(x + m + k + 1)
≥ 0,
x ≥ 0. (44)
(45)
h(0) = 0.
So, the function h(x) ≥ 0 for x ≥ 0, and we have
(n + k + 1)n+m
> 1,
(k + 1)m (n + m + k + 1)n
(46)
and
n+k+11/n n+m+k+11/(n+m)
i=k+2
i=k+2
>
n+k
i=k+1
1/n n+m+k 1/(n+m)
i
i
.
(47)
i=k+1
The proof of Theorem 2 is complete.
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INEQUALITIES AND MONOTONICITY OF SEQUENCES INVOLVING
n
(n + k)!/k!
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Department of Applied Mathematics and Informatics, Jiaozuo Institute of Technology, Jiaozuo
City, Henan 454000, P.R. China.
E-mail: [email protected], [email protected]
http://rgmia.vu.edu.au/qi.html