Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 1, Article 24, 2006
APPLICATIONS OF THE EXTENDED HERMITE-HADAMARD INEQUALITY
Z. RETKES
U NIVERSITY OF S ZEGED
B OLYAI I NSTITUTE
A RADI VÉRTANÚK TERE 1
S ZEGED , H-6720 H UNGARY
retkes@math.u-szeged.hu
Received 28 September, 2005; accepted 13 November, 2005
Communicated by P.S. Bullen
A BSTRACT. In this paper we give some combinatorial applications according to a new extension
of the classical Hermite-Hadamard inequality proved in [1].
Key words and phrases: Convexity, Hermite-Hadamard inequality, Identities.
2000 Mathematics Subject Classification. 26A51.
1. I NTRODUCTION
In the paper [1], we have proved the following generalization of the classical HermiteHadamard inequality for convex functions, extended it to n nodes: Suppose that −∞ ≤ a <
b ≤ ∞, f : (a, b) → R is a strict convex function, xi ∈ (a, b), i = 1, . . . , n such that xi 6= xj if
1 ≤ i < j ≤ n. Then the following inequality holds:
n
n
X
F (n−1) (xi )
1 X
<
f (xi ).
Πi (x1 , . . . , xn )
n! i=1
i=1
Q
In the concave case < sign is changed to >. Here Πi (x1 , . . . , xn ) := nk=1 (xi −xk ) and F (0) (s),
k6=i
(s), . . ., F
(s), . . . is a sequence of functions defined recursively by F (0) (s) = f (s) and
d
F (n) (s) = F (n−1) (s), n = 1, 2, . . .. These sequences of functions of f are known as the
ds
iterated integrals of f . As an application of this main result we have proved the following
identities:
n
n
X
X
xnk
(1.1)
=
xk ,
Π
k (x1 , . . . , xn )
k=1
k=1
F
(1)
(n−1)
(1.2)
n
X
k=1
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
292-05
xn−1
k
= 1,
Πk (x1 , . . . , xn )
2
Z. R ETKES
n
n
n
X
Y
X
1
1
n−1
= (−1)
xk
2
xk
x Πk (x1 , . . . , xn )
k=1
k=1
k=1 k
(1.3)
(xk 6= 0)
and
n
n
Y
X
1
1
n−1
= (−1)
xk
xk Πk (x1 , . . . , xn )
k=1
k=1
(1.4)
(xk 6= 0).
In this paper we apply the formulae above to obtain closed combinatorial formulae and to investigate the asymptotic behaviours of these sums. For the sake of the convenience of the reader we
collect the transformational regulations for the quantity Πk (x1 , . . . , xn ) which plays a significant role in all of the formulae. Note that all of them are simple consequences of the definition
of Πk (x1 , . . . , xn ).
a) If α ∈ R, then
Πk (α + x1 , . . . , α + xn ) = Πk (x1 , . . . , xn ),
that is, Πk is shift invariant.
b)
Πk (αx1 , . . . , αxn ) = αn−1 Πk (x1 , . . . , xn ).
c)
Πk (α − x1 , . . . , α − xn ) = (−1)n−1 Πk (x1 , . . . , xn ).
d) If xk 6= 0, k = 1, . . . , n, then
1
Πk (x1 , . . . , xn )
1
,...,
= (−1)n−1 n−2 n
.
Πk
x1
xn
xk Πj=1 xj
2. A SYMPTOTICAL F ORMULAE
a) Let 1 < x1 < · · · < xn be variables and s1 = x1 , s2 = x1 + x2 , . . ., sn = x1 + · · · + xn .
Then si 6= sj if 1 ≤ i < j ≤ n, hence formula (1.1) can be applied to s1 , . . . , sn , consequently
we have
n
n
X
X
snj
(2.1)
sj =
.
Πj (s1 , . . . , sn )
j=1
j=1
For the left-hand side of (2.1)
n
X
sj = x1 + (x1 + x2 ) + · · · + (x1 + · · · + xn ) =
j=1
n
X
j · xn−j+1 .
j=1
The definition of Πj implies
Πj (s1 , . . . , sn )
= (sj − s1 ) · · · (sj − sj−1 )(sj − sj+1 ) · · · (sj − sn )
= (x2 + · · · + xj ) · · · (xj−1 + xj ) · xj (−1)n−j xj+1 (xj+1 + xj+2 ) · · · (xj+1 + · · · + xn ).
Using the above expressions we have the following identities for all 1 < x1 < · · · < xn :
(2.2)
n
X
j=1
j · xn−j+1 =
n
X
j=1
(−1)n−j (x1 + · · · + xj )n
.
(x2 + · · · + xj ) · · · (xj−1 + xj )xj xj+1 · · · (xj+1 + · · · + xn )
J. Inequal. Pure and Appl. Math., 7(1) Art. 24, 2006
http://jipam.vu.edu.au/
A PPLICATIONS OF
THE
E XTENDED H ERMITE -H ADAMARD I NEQUALITY
3
Now, if xk → 1, k = 1, . . . , n then (2.2) gives
X
n
n
X
n+1
(−1)n−j j n
=
j=
.
2
(j − 1)!(n − j)!
j=1
j=1
Multiplying both sides by (n − 1)! leads us to
n
X
(n + 1)!
n−j n n − 1
.
(−1) j
=
j
−
1
2
j=1
b) Apply now formula (1.3) to xk = 1 + nk , k = 1, . . . , n. Hence
n
n
n X
Y
1
1
k X
n−1
= (−1)
1+
k
k
2
n k=1 (1 + n ) Πk (1 + n1 , . . . , 1 + nn )
1+ n
k=1
k=1
n−1
= (−1)
n−1
= (−1)
n
n
X
1 Y
n2
(n
+
k)
nn k=1
(n + k)2 Πk ( n1 , n2 , . . . , nn )
k=1
n
n
X
1 Y
n2
(n + k)
1
nn k=1
(n + k)2 nn−1
Πk (1, . . . , n)
k=1
n
Y
n
X
n
(n + k)2 (k − 1)!(−1)n−k (n − k)!
k=1
k=1
X
n
2n
(−1)k−1 n − 1
=n·
.
n k=1 (n + k)2 k − 1
= (−1)n−1
(n + k)
From this equality chain we conclude that
X
n
n
(−1)k−1 n − 1
2n
1X 1
(2.3)
=
n k=1 (n + k)2 k − 1
n k=1 1 +
k
n
=
n
X
k=1
1
.
n+k
Since
n
1X 1
lim
n→∞ n
1+
k=1
Z
k
n
=
0
1
1
du = log 2,
1+u
hence the left-hand side of (2.3) is convergent and
X
n
2n
(−1)k−1 n − 1
lim
= log 2.
n→∞
n k=1 (n + k)2 k − 1
n
Pursuant to the Stirling formula 2n
∼ √4πn (n → ∞), consequently we have the following
n
asymptotic expression for the sum on the left:
√
n
X
√
(−1)k−1 n − 1
n
∼ π log 2 n (n → ∞).
2
(n + k) k − 1
4
k=1
2
c) This example follows along the lines above but choosing xk = 1 + nk (k = 1, . . . , n).
Without the detailed computation we have
Qn
n
n
n
2
2 X
k−1 2
(n
+
k
)
1X
1
(−1)
k
(2.4)
= 2n k=1 2
· k .
2 + k 2 )2 n+k
n k=1 1 + ( nk )2
(n!)
(n
n
k=1
J. Inequal. Pure and Appl. Math., 7(1) Art. 24, 2006
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4
Z. R ETKES
From this equality the limit of the right-hand side exists and
Qn
n
n
n
2
2 X
(−1)k−1 k 2 k
1
1X
k=1 (n + k )
lim 2n ·
· n+k = lim
2
2
2
2
n→∞
n→∞
(n!)
(n + k )
n k=1 1 + ( nk )2
n
k=1
Z 1
1
du
=
2
0 1+u
π
= arctan 1 = .
4
Combining these observations we have
n
n
X
(−1)k−1 k 2 k
π
(n!)2
∼
·
(n → ∞).
(n2 + k 2 )2 n+k
8n Πnk=1 (n2 + k 2 )
n
k=1
Moreover (n!)2 ∼ 2πn( ne )2n and
n
Y
(n2 + k 2 ) = n2n
k=1
n
Y
"
k=1
2 #
k
1+
.
n
On the other hand
"
"
2 #
2 #
Z 1
n
n
Y
1
k
k
1X
log
1+
=
log 1 +
→
log(1 + x2 )dx
n
n
n k=1
n
0
k=1
(n → ∞),
hence
n
Y
k=1
"
2 #
R1
k
2
1+
∼ en 0 log(1+x )dx .
n
Using integration by parts to evaluate the integral in the exponent gives
Z 1
π
log(1 + x2 )dx = log 2 + − 2 = α > 0.
2
0
Putting together these results gives the following asymptotical expression:
n
n
X
(−1)k−1 k 2 k
π 2 −n(log 2+ π )
2
∼
e
(n → ∞).
2 + k 2 )2 n+k
(n
4
n
k=1
3. F ORMULAE FOR P OWER S UMS
Let u1 , . . . , un ∈ R such that ui 6= 0, |ui | < 1, ui 6= uj if 1 ≤ i < j ≤ n and xi = 1 − ui
(i = 1, . . . , n). Then applying (1.3) gives
n
X
i=1
n
n
Y
X
1
1
n−1
= (−1)
(1 − uk )
1 − ui
(1 − ui )2 Πi (1 − u1 , . . . , 1 − un )
i=1
k=1
=
n
Y
(1 − uk )
i=1
k=1
Under the above conditions
(3.1)
n
X
i=1
1
1−ui
n
X
=
P∞
`=0
n
1
.
(1 − ui )2 Πi (u1 , . . . , un )
u`i , hence
∞
∞
n
∞
XX
XX
X
1
=
u`i =
u`i =
P (`),
1 − ui
i=1 `=0
`=0 i=1
`=0
J. Inequal. Pure and Appl. Math., 7(1) Art. 24, 2006
http://jipam.vu.edu.au/
A PPLICATIONS OF
THE
E XTENDED H ERMITE -H ADAMARD I NEQUALITY
5
where we use the denotation P (`) := u`1 + · · · + u`n for (` = 0, 1, 2, . . .), and thus we have the
following identities for u1 , . . . , un :
(3.2)
∞
X
P (`) =
`=0
Apply now (3.2) to uk =
n
Y
n
X
(1 − uk )
i=1
k=1
(1 − ui
)2 Π
1
.
i (u1 , . . . , un )
k
n+1
(k = 1, . . . , n). For the left-hand side:
` X
∞
n
∞
∞
∞ X
n X
X
X
X
1
P̃ (`)
k
`
P (`) =
=
k
=
,
`
`
n
+
1
(n
+
1)
(n
+
1)
`=0
k=1
`=0
`=0
`=0 k=1
where P̃ (`) = 1` + 2` + · · · + n` (` = 0, 1, 2, . . .). On the other hand
n Y
k
n!
1−
=
and
n
n
+
1
(n
+
1)
k=1
1
n
(−1)n−i (i − 1)!(n − i)!
Πi
,...,
=
.
n+1
n+1
(n + 1)n−1
Putting these expressions into the right-hand side of (3.2) we obtain
∞
n
X
X
P̃ (`)
(−1)j−1 n − 1
(3.3)
= n(n + 1)
.
(n + 1)`
j2
j−1
j=1
`=0
= nj gives
A simple rearrangemant of (3.3) and using nj n−1
j−1
∞
n
X
X
P̃ (`)
(−1)j−1 n
=
,
`+1
(n
+
1)
j
j
j=1
`=0
and since it is known that
n
X
(−1)j−1 n
j=1
j
j
Z
=
0
1
n
X1
1 − (1 − s)n
ds =
,
s
k
k=1
consequently we have the following identity for all fixed n:
∞
X
`=0
n
X1
P̃ (`)
=
.
(n + 1)`+1
k
k=1
Let us start now with (3.1) substituting uk = x1k , xk > 1 (k = 1, . . . , n) and xi 6= xj if
1 ≤ i < j ≤ n. Using the same technique applied above and the transformational rule d) we
get the following identity for x1 , . . . , xn :
(3.4)
∞ X
n
n
n
Y
X
X
1
xni
n−1
=
(−1)
(x
−
1)
.
k
`
2 Π (x , . . . , x )
(x
−
1)
x
i
i
1
n
k
i=1
`=0 k=1
k=1
Putting xk = k + 1 (k = 1, . . . , n) in (3.4) and simplifying the right-hand side we have the
following equality:
X
∞ n
X
1
1
(−1)i−1
n n
(3.5)
+ ··· +
=
(i + 1)
.
`
`
2
(n
+
1)
i
i
i=1
`=0
1
[n+1]
Observe that 21` + · · · + (n+1)
(`) − 1, where ζ [n+1] (`) denotes the (n + 1)th partial
` = ζ
P∞ 1
sum of the series ζ(`) = k=1 k` , if ` ≥ 2. Moreover let Hn denote the nth partial sum of
J. Inequal. Pure and Appl. Math., 7(1) Art. 24, 2006
http://jipam.vu.edu.au/
6
Z. R ETKES
P
the harmonic series, that is Hn = nk=1 k1 . Using this quantity and separating the summands
corresponding to ` = 0 and ` = 1 gives
(3.6)
n + Hn+1 − 1 +
∞
X
[ζ
[n+1]
(`) − 1] =
n
X
(−1)i−1
i
i=1
`=2
n
(i + 1)
.
i
n
Rearranging (3.6) and taking the limit as n → ∞ yields that:
#
" n
∞
X
X (−1)i−1
n
(3.7)
[ζ(`) − 1] = 1 + lim
(i + 1)n
− n − Hn+1 .
n→∞
i
i
1=1
`=2
This relation prompted us to investigate the quantities arising in both sides of the equality.
On one hand the sum of the series is equal to 1 as an easy computation shows below:
∞
X
∞ X
∞
∞
∞
X
X
1
1 X 1
[ζ(`) − 1] =
=
k`
k 2 j=0 k j
`=2
`=2 k=2
k=2
∞
X
1 1
=
k2 1 −
k=2
=
∞
X
k=2
1
k
1
= 1.
k(k − 1)
Hence
"
lim
n→∞
n
X
(−1)i−1
i=1
i
#
n
(i + 1)n
− n − Hn+1 = 0.
i
In fact we prove more in the following lemma.
Lemma 3.1. For all n ≥ 1 the identity below holds
n
X
(−1)i−1
i=1
i
n
(i + 1)
= n + Hn .
i
n
Proof. In the paper [1] we proved that if x1 , . . . , xn ∈ R such that xi 6= xj if 1 ≤ i < j ≤ n
then
 Pn
if j = n

k=1 xk

n
j

X
xi
1
if j = n − 1
=
Πi (xi , . . . , xn ) 

i=1

0
if 0 ≤ j ≤ n − 2.
Applying this result to xi =
1
i
(i = 1, . . . , n) gives that

Hn


n

X
(−1)i−1 n−j n
1
i
=

i
i

i=1

0
J. Inequal. Pure and Appl. Math., 7(1) Art. 24, 2006
if
j=n
if
j =n−1
if
0 ≤ j ≤ n − 2.
http://jipam.vu.edu.au/
A PPLICATIONS OF
THE
E XTENDED H ERMITE -H ADAMARD I NEQUALITY
7
Moreover
n
X
(−1)i−1
i=1
i
" n
# X
n
i−1 X
n
(−1)
n
n
in−k
(i + 1)n
=
i
i
k
i
i=1
k=0
"
#
n
n
X
X
(−1)i−1 n−k n
n
=
i
i
i
k
k=0 i=1
n
n
= n + Hn
=
+ Hn
n−1
n
as we stated.
R EFERENCES
[1] Z. RETKES, An extension of the Hermite-Hadamard inequality (submitted)
[2] S.S. DRAGOMIR AND C.E.M. PEARCE, Selected Topics on Hermite-Hadamard Inequalities,
RGMIA Monographs, Victoria University, 2000. [ONLINE: http://rgmia.vu.edu.au/
monographs/index.html]
[3] W. RUDIN, Real and Complex Analysis, McGraw-Hill Book Co., 1966.
[4] G.H. HARDY, J.E. LITTLEWOOD
1934.
J. Inequal. Pure and Appl. Math., 7(1) Art. 24, 2006
AND
G. PÓLYA, Inequalities, Cambridge University Press,
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