Applied Mathematics E-Notes, 15(2015), 187-196 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
The Evaluation Of A Quadratic And A Cubic Series
With Trigamma Function
Ovidiu Furduiy
Received 26 May 2015
Abstract
The paper is about calculating the quadratic series
1
X
1
n
n=1
(2)
1
1
22
1
n2
2
=
1
X
1
(
n
n=1
0
(n + 1))2
and the cubic series
1
X
n
n=1
where
1
1
1
+
+
n2
(n + 1)2
3
=
1
X
n(
0
(n))3 ;
n=1
denotes the digamma function.
Introduction and the Main Result
Throughout this paper, let C, Z0 , N denote the sets of complex numbers, nonpositive
integers, positive integers respectively. The celebrated Riemann zeta function is a
function of a complex variable [9, p.265] de…ned by
(z) =
1
X
1
1
1
=1+ z + z +
z
n
2
3
n=1
+
1
+
nz
(<(z) > 1) :
When z = 2 one has that the Riemann zeta function value (2) is de…ned by the series
formula
1
X
1
1
1
1
(2) =
=1+ 2 + 2 +
+ 2+
:
2
n
2
3
n
n=1
The trigamma function
0
(z) =
0
is de…ned by [7, p.22]
d2
d
log (z) =
(z)
dz 2
dz
z 2 C n Z0 ;
Mathematics Subject Classi…cations: 11M06, 33B15, 33E20, 40A05.
of Mathematics, Technical University of Cluj-Napoca, Str. Memorandumului Nr. 28,
400114, Cluj-Napoca, Romania
y Department
187
188
The Evaluation of a Quadratic and a Cubic Series
0
(z)
d
(z) being the
(or digamma) function de…ned by (z) = dz
log (z) = (z)
or,
in terms of the generalized (or Hurwitz) zeta function (s; a) de…ned by (s; a) :=
1
P
1
<(s) > 1; a 2 C n Z0 ,
(k+a)s
k=0
0
(z) =
1
X
k=0
1
= (2; z)
(k + z)2
z 2 C n Z0 :
This implies that
0
(n) =
1
1
+
+
n2
(n + 1)2
= (2)
1
22
1
1
(n
1)2
(n 2 N n f1g) :
Closed form evaluation of series involving (k) are collected in [7] and, more recently,
in [8]. Other series, linear or quadratic, involving the Riemann zeta function and
harmonic numbers, which are evaluated in terms of special constants can be found in
[4].
In this paper we evaluate a quadratic and a cubic series involving the tail of (2).
More precisely, we calculate the quadratic series
1
X
1
n
n=1
(2)
1
1
22
2
1
n2
1
X
1 0
( (n + 1))2
=
n
n=1
and the cubic series
1
X
1
1
+
+
n2
(n + 1)2
n
n=1
3
=
1
X
n( 0 (n))3 ;
n=1
where denotes the digamma function.
The main result of this paper is the following theorem.
THEOREM 1 (A quadratic and a cubic series with the tail of (2)). The following
identities hold:
(a)
(b)
P1
1
n=1 n
P1
n=1
n
(2)
0
(n)
1
3
=
1 2
n2
1
22
9
2
(3)
17
8
= 5 (2) (3)
(4)
25
4
(5) +
9
2
9 (5);
(2) (3).
We need in our analysis Abel’s summation formula [1, p.55], [4, p.258] which states
that
Pn if (an )n 1 and (bn )n 1 are two sequences of real or complex numbers and An =
k=1 ak , then
n
X
k=1
ak bk = An bn+1 +
n
X
k=1
Ak (bk
bk+1 )
(n 2 N) :
O. Furdui
189
We will also use, in our calculations, the in…nite version of the preceding formula:
1
X
ak bk = lim (An bn+1 ) +
n!1
k=1
1
X
Ak (bk
bk+1 );
(1)
k=1
provided the in…nite series on the right hand side of (1) converges and the limit is …nite.
A special function which is used in the proof of part (a) of Theorem 1 is the
Dilogarithm function. Recall that the Dilogarithm function Li2 is de…ned, for jzj 1,
by [7, p.106]
Z z
1
X
ln(1 t)
zn
=
dt:
Li2 (z) =
2
n
t
0
n=1
In particular, Li2 (1) = (2).
A special identity involving the Dilogarithm function is the following Landen type
formula:
(2) Li2 (1 z) ln z ln(1 z) = Li2 (z);
(2)
whose proof can be found in [7, p.107].
2
Proof of the Main Result
In this section we collect some results we need for proving Theorem 1.
LEMMA 1 (Some logarithm and polylogarithm integrals). The following equalities
hold:
(a)
(b)
(c)
(d)
(e)
R1
0
R1
0
R1
0
R1
0
R1
0
x ln x
1 x dx
=1
ln x ln(1 x)
dx
x
(2);
= (3);
Li2 (x)
x dx
= (3);
Li22 (x)
x dx
= 2 (2) (3)
ln x ln(1 x)Li2 (x)
dx
x
3 (5);
= (2) (3)
3
2
(5).
PROOF. (a) We have
Z
0
1
x ln x
dx
1 x
=
Z
1
x ln x
0
=
1
X
n=0
1
X
x
n
!
1
=1
(n
+
2)2
n=0
dx =
1 Z
X
n=0
(2):
0
1
xn+1 ln xdx
190
The Evaluation of a Quadratic and a Cubic Series
(b) We have
Z
1
0
ln x ln(1
x
x)
dx
Z
=
1
0
=
1
X
xn
n
n=1
ln x
x
Z
1
X
1 1 n
x
n 0
n=1
1
!
dx
ln xdx = (3):
(c) We have
Z
0
1
Z
Li2 (x)
dx =
x
1
0
1
x
1
X
xn
n2
n=1
!
dx =
Z 1
1
X
1
xn
2
n
0
n=1
1
dx = (3):
The integrals in parts (d) and (e) are recorded in [3, Entry 1, Table 2, p.1435, Entry
2, Table 6, p.1436].
The next lemma is about calculating two Euler series and a quadratic series involving the tail of (2).
LEMMA 2. The following equalities hold:
P1 1
1
(a)
+ n12 = 92 (5) + 3 (2) (3);
n=1 n3 1 + 22 +
P1 1
1
(b)
+ n12 = 74 (4);
n=1 n2 1 + 22 +
(c)
P1
n=1
(2)
1
1 2
n2
1
22
5
2
= 3 (3)
(4).
PROOF. (a) This part of the lemma is a special case of a more general result
concerning the evaluation of Euler type series [2, Theorem 3.1, p.22].
(b) We apply Abel’s summation formula (1) with an = n12 and bn = 1+ 212 + + n12 .
We have
s =
=
1
X
1
n2
n=1
lim
n!1
1
X
1+
1
+
22
1+
1+
1
+
22
+
1
(n
+
1)2
n=1
1+
1
+
22
1
(n + 1)2
n=1
=
2
=
2
=
(2)
1
n2
1
n2
1+
1
+
22
+
1
X
+
(2) s + 1 + (4)
7
(4) s;
2
1
1
+
22
+
1
(n + 1)2
1
n2
+
1
(n + 1)2
+
1
X
1
(n
+
1)4
n=1
O. Furdui
191
and part (b) of the lemma is proved.
2
4
We used that 2 (2) = 52 (4) since (2) = 6 and (4) = 90 [6, p.605].
(c) The evaluation of this quadratic series involving the tail of (2) can be found in
[4, Problem 3.22, p.142], [5, Theorem 1, (a)].
Now we are ready to prove Theorem 1.
PROOF. (a) First we note that if k > 0 is a real number then
Z
1
xk
1
1
;
k2
ln x dx =
0
and this implies that
(2)
1
1
22
1
n2
=
=
1
X
1
(n
+
m)2
m=1
Z
1
1
X
xn+m
m=1
Z
=
1
X
1
n
x ln x
0
Z
=
1
ln x dx
0
x
m=1
1
0
xn
1
x
m 1
!
dx
ln x dx:
It follows that
1
X
1
n
n=1
T
(2)
1
1
22
1
n2
2
Z 1 n
Z 1 n
1
X
1
y
x
=
ln x dx
ln y dy
n
1
x
1
y
0
0
n=1
Z 1Z 1
Z 1Z 1
1
X
(xy)n
ln x ln y
ln x ln y ln(1 xy)
=
dxdy =
dxdy:
x)(1 y) n=1 n
(1 x)(1 y)
0
0
0 (1
0
We have
I=
Z
1
0
Z
0
1
ln x ln y ln(1 xy)
dxdy =
(1 x)(1 y)
Z
0
1
ln x
1 x
Z
1
0
ln y ln(1 xy)
dy dx:
1 y
We calculate the inner integral by parts, with f (y) = ln(1
0
g (y) =
Z
0
1
ln y
1 y,
g(y) =
ln y ln(1 xy)
dy
1 y
ln y ln(1
y)
xy), f 0 (y) =
Li2 (y), and we have
y=1
=
ln(1
xy)(ln y ln(1
y) + Li2 (y))
y=0
x
1 xy ,
192
The Evaluation of a Quadratic and a Cubic Series
Z
0
=
1
x
1
(ln y ln(1
xy
(2) ln(1
Z
x)
y) + Li2 (y)) dy
1
x
1
0
(ln y ln(1
xy
y) + Li2 (y)) dy:
It follows, based on part (b) of Lemma 1, that
I
=
(2)
Z
1
Z
1
ln x ln(1 x)
dx
1 x
0
1
0
0
=
Z
(1
(2) (3)
x ln x
(ln y ln(1 y) + Li2 (y)) dxdy
x)(1 xy)
Z 1Z 1
x ln x
(ln y ln(1 y) + Li2 (y)) dxdy:
(1
x)(1
xy)
0
0
We calculate the double integral as follows
J
=
=
Z
1
Z
1
x ln x
(ln y ln(1
x)(1
xy)
0
0
Z 1
Z 1
(ln y ln(1 y) + Li2 (y))
(1
0
0
y) + Li2 (y)) dxdy
x ln x
dx dy:
x)(1 xy)
(1
Using part (a) of Lemma 1 the inner integral becomes
Z
0
1
x ln x
dx
x)(1 xy)
(1
=
Z
=
1
0
ln x
dx
1 xy
1
1
1
=
Z
x ln x
1 y
Z 1
0
=
Using the substitution 1
1
y
0
1
1
y
x
x ln x
dx
1 x
1
xy
1
dx
Z
1
xy ln x
dx
xy
0 1
Z 1
Z 1
ln x
dx
ln xdx
1
xy
0
0
1
y
1
(2)
+
1 y
1 y
Z 1
(2)
1
ln x
dx:
1 y 1 y 0 1 xy
xy = t; we get that
=
=
1
y
1
y
Z
1 y
ln(1
1
Z
1 y
1
t)
t
ln y
ln(1 t)
dt
t
dt
ln y ln(1
y) :
On the other hand,
Z
1
1 y
ln(1 t)
dt
t
=
=
Z
Z 1
ln(1 t)
ln(1 t)
dt
dt
t
t
0
0
Li2 (1 y) + Li2 (1)
1 y
O. Furdui
193
=
(2)
Li2 (1
y);
and it follows, based on formula (2), that
Z
1
0
1
ln x
dx =
xy
Therefore
Z
1
( (2)
y
1
(1
0
Li2 (1
y)
x ln x
dx =
x)(1 xy)
and this in turn implies that
Z 1
(ln y ln(1
J =
ln y ln(1
1
0
=
Z
1
=
Z
1
0
+
Z
1
(ln y ln(1
y) + Li2 (y))
Using Lemma 1 combined to ln y ln(1
1
(ln y ln(1
y) + Li2 (y))
=
1
( (2)
Li2 (1
( (2)
Li2 (y)) (Li2 (1
y
0
=
Z
1
0
Z
1
dy
Li2 (y)
(2)
dy:
1 y
y) + Li2 (y) = (2)
0
Z
dy
y) + Li2 (y))
0
Z
(2)
1 y
(2)
Li2 (y) Li2 (y)
+
y
1 y
Z 1 2
Li2 (y)
ln y ln(1 y)Li2 (y)
dy +
dy
y
y
0
(ln y ln(1
0
(2)
Li2 (y)
+
;
y y(1 y)
Li2 (y)
y(1 y)
y) + Li2 (y))
Li2 (y)
:
y
y)) =
(3)
Li2 (1
y), we have that
Li2 (y)
(2)
dy
1 y
y)) (Li2 (y)
1 y
(2))
y)
(2))
dy
(y ! 1
y)
dy
Li2 (y)) ( Li2 (y) ln y ln(1 y))
dy by (2)
y
0
Z 1
Z 1
Li2 (y)
ln y ln(1 y)
=
(2)
dy
(2)
dy
y
y
0
0
Z 1 2
Z 1
Li2 (y)
Li2 (y) ln y ln(1 y)
+
dy +
dy
y
y
0
0
Z 1 2
Z 1
Li2 (y)
Li2 (y) ln y ln(1 y)
=
2 (2) (3) +
dy +
dy:
y
y
0
0
=
( (2)
We obtain in view of (3), (4) and parts (d) and (e) of Lemma 1 that
J
=
2
Z
1
ln y ln(1
0
=
4 (2) (3)
y)Li2 (y)
y
9 (5);
dy + 2
Z
0
1
Li22 (y)
dy
y
2 (2) (3)
(4)
194
The Evaluation of a Quadratic and a Cubic Series
and hence
I=
(2) (3)
J = 9 (5)
5 (2) (3):
Since T = I we get that part (a) of the theorem is proved.
(b) We apply Abel’s summation formula (1) with an = n and bn = x3n where
0
xn =
(n) =
1
1
+
+
2
n
(n + 1)2
+
:
A calculation shows that
bn
bn+1 =
1
+ xn+1
n2
3
1
3
3
+ 4 xn+1 + 2 x2n+1 ;
n6
n
n
x3n+1 =
and we have
1
X
n=1
n
3
1
1
+
+
n2
(n + 1)2
=
n(n + 1)
n!1
2
1
1
+
+
(n + 1)2
(n + 2)2
lim
1
1X
n(n + 1)
+
2 n=1
=
1
3
3
+ 4 xn+1 + 2 x2n+1
n6
n
n
1
1
1
1
3 X xn+1
3 X xn+1
(4) + (5) +
+
2
2
2 n=1 n2
2 n=1 n3
+
1
1
3 X x2n+1
3X 2
xn+1 +
:
2 n=1
2 n=1 n
(5)
The preceding limit is 0 since
<
3
n(n + 1)
1
1
+
+
2
(n + 1)
(n + 2)2
n(n + 1)
1
1
+
+
n(n + 1) (n + 1)(n + 2)
3
<
and the limit follows based on the Squeeze Theorem.
Since
1
xn+1 = 0 (n + 1) = (2) 1
22
we have, based on parts (a) and (b) of Lemma 2, that
1
X
xn+1
n2
n=1
=
1
X
(2)
1
n=1
=
=
2
(2)
5
(4)
2
3
1
X
1
2
n
n=1
7
(4)
4
1
22
n2
1+
1
;
n2
1
n2
1
+
22
+
1
n2
n+1
;
n2
O. Furdui
195
3
(4)
4
=
(6)
and
1
X
xn+1
n3
n=1
=
1
X
(2)
n=1
=
(2) (3)
=
(2) (3)
=
1
1
22
n3
1
X
1
n3
n=1
1
n2
1+
1
+
22
+
1
n2
9
(5) + 3 (2) (3)
2
9
2 (2) (3) + (5):
2
(7)
Combining (5), (6), (7), part (c) of Lemma 2 and part (a) of Theorem 1 we have
1
X
n=1
n
1
1
+
+
n2
(n + 1)2
3
=
9
(3)
2
17
(4)
8
25
9
(5) + (2) (3);
4
2
and the theorem is proved.
A challenging problem would be to evaluate the alternating versions of the series in
Theorem 1. We leave this as an open problem to the interested reader.
Acknowledgment. The author thanks Alina Sînt¼
am¼
arian for suggesting the problem of evaluating the cubic series in the second part of Theorem 1.
References
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am¼
arian, Quadratic series involving the tail of (k), Integral
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196
The Evaluation of a Quadratic and a Cubic Series
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