SOME INTERESTING SERIES ARISING FROM THE POWER
SERIES EXPANSION OF (sin−1 x)q
HABIB MUZAFFAR
Received 3 February 2005 and in revised form 23 June 2005
Starting from the power series expansions of (sin−1 x)q , for 1 ≤ q ≤ 4, formulae are obtained for the sum of several infinite series. Some of these evaluations involve ζ(3).
1. Introduction
In [10], Choe deduced the formula
∞
1
n =1
n2
=
π2
6
(1.1)
from the power series expansion of sin−1 (x) (see also [1, 16]). By applying a generalization of the procedure used by Choe to the power series expansions of (sin−1 x)q for
1 ≤ q ≤ 4, we obtain explicit formulae for the sum of several infinite series, see (2.1),
(2.2), (2.3), (2.4), (2.5), and (2.6). For other applications based on the procedure used by
Choe, see [11, 12, 17].
2. Main results
Let m denote an integer. For m ≥ 0, we have the following theorems.
Theorem 2.1.
k =0

2k
k
∞
(2k + 1)(2k + 2m + 1)

= 2−4m 

2k+2m
k+m
m
r =1
r ≡1( mod 2)
2m
m−r
r2
+
2m
m

π2 

.
8
(2.1)
Theorem 2.2.
2k+2m
k+m
2k
2
k =1 k
k
∞
=
m 2
r =1
2m
m−r
r2
2m π 2
+
.
m 6
Copyright © 2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:14 (2005) 2329–2336
DOI: 10.1155/IJMMS.2005.2329
(2.2)
2330
Series arising from power series of (sin−1 x)q
Theorem 2.3.
2k
k
∞
k =1
(2k + 1)(2k + 2m + 1)

=2

−
1
(2
j
−
1)2
j =1
2k+2m
k+m
m
−4m−1 
k
r =1
r ≡1( mod 2)
2m
m−r
2r 4
+π
2
m
r =1
2m
m−r
8r 2

2m π 4 
.
+
m 192 
(2.3)
Theorem 2.4.
2k+2m+2
k+m+1
∞
k =1
(k + 1)(2k + 1)
2k
k
k
1
j =1
j2
= −4
m
r =1
2m
m−r
r4
2m
2π 2 m−r
2m π 4
+
+
.
2
m 60
3 r =1 r
m
(2.4)
In addition, we have the following theorems.
Theorem 2.5.
∞
1
1
π2
7
=
log 2 − ζ(3),
2
k(2k
+
1)
(2
j
−
1)
4
8
j =1
k =1
k
∞
(2.5)
1
1
π2
3
=
log 2 − ζ(3).
2
(k + 1)(2k + 1) j =1 j
3
2
k =1
k
Theorem 2.6.
∞
1
k
π2 2
1
π2
=−
+ log2 + log 2 − ζ(3).
2
(k
+
1)(2k
+
1)(2k
−
1)
j
36
3
9
2
j =1
k =1
k
(2.6)
In (2.5) and (2.6), ζ represents the Riemann zeta function.
The following result in [14] (m ≥ 0) should be compared with (2.1) :
∞
k =0
2k
k
(2k + 2m + 1)(2k + 4m + 1)
=
2k+4m
k+2m
π2
28m+3
2m
m
2
.
(2.7)
Also, the series appearing above in (2.3), (2.4), (2.5), and (2.6) bear some resemblance
to Euler sums (see, e.g., [3, 4, 5, 9]). A very broad generalization which generalizes both
Euler sums and polylogarithms is studied in [6]. For other interesting evaluations of series
involving binomial coefficients, see, for example, [7, 8, 15, 18].
Habib Muzaffar
2331
3. Proofs of Theorems 2.1, 2.2, 2.3, and 2.4
The power series expansions of (sin−1 x)q for 1 ≤ q ≤ 4 (valid for |x| ≤ 1) are given by
(see [10], [2, pages 262-263])
−1
sin x =
∞
k =0
sin−1 x
2
=
2k
k
22k
x2k+1
,
2k + 1
∞
22k−1 x2k
,
2
2k
k =1
−1
sin x
3
=6
∞
k =1
−1
sin x
4

2k
k
22k
k
k

k


∞
k
22k  1 
=3
2k
k
k =1
(3.1)
2k+1

 x
,
(2 j − 1)2 2k + 1
j =1
1
j =1
j2
x2k+2
.
(k + 1)(2k + 1)
Multiplying each of (3.1) by x2m , where m is an integer, putting x = sinθ and integrating
with respect to θ from θ = 0 to θ = π/2, and using the well-known results (valid for
nonnegative integers p)
π/2
0
sin2p+1 θ dθ =
π/2
0
22p
,
(2p + 1) 2p
p
sin2p θ dθ =
2p
p
2p
2
(3.2)
π
,
2
we obtain
π/2
0
2m
θ sin
θ dθ = 2
k =0
π/2
0
2
θ sin
0
π/2
2m
θ 3 sin2m θ dθ = 3 2
2m
0
4
2m
θ sin
2k
k
2k+2m
k+m
22m+2 k=1 k2 2k
k
π
∞
,
2k+2m
k+m
2k
k
(2k + 1)(2k + 2m + 1)
∞
3π 2k+2m+2
k+m+1
2k+2m
k+m
22m+3 k=1 (k + 1)(2k + 1)
2k
k
,
m ≥ 0,
m ≥ −1,
∞
2m+1
θ dθ =
(2k + 1)(2k + 2m + 1)
θ dθ =
k =1
π/2
∞
(3.4)
k
1
j =1
(2 j − 1)2
,
m ≥ −1,
(3.5)
k
1
j =1
(3.3)
j2
,
m ≥ −2.
(3.6)
2332
Series arising from power series of (sin−1 x)q
For m ≥ 0, we evaluate the integrals on the left of (3.3), (3.4), (3.5), and (3.6) using the
following formula valid for a nonnegative integer m (see [13, page 31]):
2m
sin
θ=2
−2m

−1
m

(−1)
m+ j
j =0

2m
2m 
2
cos 2(m − j)θ +
,
j
m 
(3.7)
and the following easily checked formulae (valid for positive integers l):
π/2
(−1)l − 1
,
4l2
0
π/2
(−1)l π
θ 2 cos(2lθ)dθ =
,
4l2
0
π/2
(−1)l π 2 1 − (−1)l
3
θ cos(2lθ)dθ = 3
+
,
16l2
8l4
0
π/2
0
θ cos(2lθ)dθ =
θ 4 cos(2lθ)dθ = (−1)l π
(3.8)
π2
3
−
.
8l2 4l4
After some simplification, we obtain (2.1), (2.2), (2.3), and (2.4).
4. Special cases of Theorems 2.1, 2.2, 2.3, and 2.4
We record the special cases corresponding to 0 ≤ m ≤ 2.
Putting m = 0, 1, 2 in (2.1), we get
∞
1
π2
=
,
2
(2k + 1)
8
k =0
∞
k+1
1 π2
= +
,
2
(2k + 1) (2k + 3) 8 32
k =0
2k
k
∞
k =0
(4.1)
(2k + 1)(2k + 5)
=
2k+4
k+2
1
3π 2
+
.
64 1024
Putting m = 0,1,2 in (2.2), we get
∞
1
k =1
k2
=
π2
,
6
2k+2
k+1
2k
2
k =1 k
k
= 2+
2k+4
k+2
2 2k
k
k =1
k
=
∞
∞
π2
,
3
17
+ π2.
2
The first results of (4.1) and (4.2) are of course well-known classical results.
(4.2)
Habib Muzaffar
2333
Putting m = 0,1,2 in (2.3), we get
∞
k
1
1
π4
=
,
(2k + 1)2 j =1 (2 j − 1)2 384
k =1
2k
k
∞
k =1
(2k + 1)(2k + 3)
2k+2
k+1
2k
k
∞
k =1
(2k + 1)(2k + 5)
2k+4
k+2
k
1
j =1
(2 j − 1)2
k
1
j =1
(2 j − 1)2
=
=
−1
64
−1
256
+
π2
π4
+
,
256 3072
+
17π 2
π4
+
.
16384 16384
(4.3)
Putting m = 0,1,2 in (2.4) gives
∞
∞
k =1
2k+4
k+2
(k + 1)(2k + 1)
∞
k =1
k
1
1
π4
=
,
2
2
(k + 1) j =1 j
120
k =1
2k+6
k+3
(k + 1)(2k + 1)
2k
k
2k
k
k
1
j =1
j2
k
1
j =1
j2
2π 2 π 4
+ ,
3
30
= −4 +
=−
(4.4)
65 17π 2 π 4
+
+ .
4
6
10
We note that the first series evaluated in (4.4) is an Euler sum and the result is classical
and was known to Euler (see, e.g., [5]).
5. Proof of Theorem 2.5
We consider the case m = −1 of (3.5), (3.6) (the case m = −1 of (3.4) gives a trivial result).
We need the following result valid for a positive integer n and |x| < 2π (see [2, page 260]):
x
0
un
u
nπ
Γ(n + 1) n− j
du = cos
n!ζ(n + 1) − (−1) j( j+1)/2
cot
x Cl j+1 (x),
2
2
2
Γ(n + 1 − j)
j =0
(5.1)
n
where
Cl2n (x) =
∞
sin(kx)
k =1
Cl2n+1 (x) =
k2n
∞
cos(kx)
k =1
k2n+1
,
(5.2)
,
Series arising from power series of (sin−1 x)q
2334
and Γ and ζ represent the Gamma function and the Riemann zeta function respectively.
We note that
Cl2n (π) = 0,
1
Cl2n+1 (π) = 2n − 1 ζ(2n + 1),
2
Cl1 (π) = − log2.
n ≥ 1,
(5.3)
Putting x = π in (5.1), we obtain
2n
π/2
0
θ n cotθ dθ = n!cos
nπ
Γ(n + 1) n− j
ζ(n + 1) − (−1) j( j+1)/2
π Cl j+1 (π).
2
Γ(n
+ 1 − j)
j =0
n
(5.4)
Using
π/2
0
θ n cotθ dθ =
1
n+1
π/2
0
θ n+1 csc2 θ dθ,
n ≥ 1,
(5.5)
in (5.4), we get
2n
n+1
π/2
0
θ n+1 csc2 θ dθ
= n!cos
nπ
Γ(n + 1) n− j
ζ(n + 1) − (−1) j( j+1)/2
π Cl j+1 (π).
2
Γ(n + 1 − j)
j =0
n
(5.6)
From (5.6) and (5.3) we obtain
π/2
0
π/2
θ 2 csc2 θ dθ = π log2,
(5.7)
3
21
θ 3 csc2 θ dθ = π 2 log2 − ζ(3),
4
8
0
π/2
3
π
9
θ 4 csc2 θ dθ =
log2 − πζ(3).
2
4
0
(5.8)
(5.9)
Putting m = −1 in (3.5) and (3.6) and using (5.8) and (5.9) give (2.5).
6. Proof of Theorem 2.6
We consider the case m = −2 of (3.6). We need to evaluate
π/2
0
4
4
4
2
θ csc θ dθ = θ csc θ(− cotθ)
=4
π/2
0
π/2
0
π/2
0
π/2
+
cotθ
0
θ 3 cotθ csc2 θ dθ − 2
π/2
0
θ 4 csc4 θ dθ. We have
d 4 2 θ csc θ dθ
dθ
θ 4 csc2 θ cot2 θ dθ.
(6.1)
Habib Muzaffar
2335
Using cot2 θ = csc2 θ − 1 in the second integral on the right gives
π/2
0
θ 4 csc4 θ dθ =
4
3
π/2
0
θ 3 cotθ csc2 θ dθ +
2
3
π/2
0
θ 4 csc2 θ dθ.
(6.2)
Also,
π/2
0
θ 3 cotθ csc2 θ dθ = θ 3 cscθ(− cscθ)
π3
=−
+3
8
π/2
0
2
π/2
0
π/2
+
cscθ
0
2
θ csc θ dθ −
π/2
0
d 3
θ csc θ dθ
dθ
3
(6.3)
2
θ cotθ csc θ dθ,
so that
π/2
0
θ 3 cotθ csc2 θ dθ = −
π3 3
+
16 2
π/2
0
θ 2 csc2 θ dθ.
(6.4)
From (6.2), (6.4), (5.7), and (5.9), we obtain
π/2
0
θ 4 csc4 θ dθ = −
π3
3
π3
+ 2π log2 + log2 − πζ(3).
12
3
2
(6.5)
Putting m = −2 in (3.6) and using (6.5), we obtain (2.6).
7. Final remarks
In a future paper, we plan to investigate what happens when we multiply (3.1) by x2m+1
and carry out the same steps as we did here.
Acknowledgments
The research for this paper was carried out while the author was a Postdoctoral Fellow at
the University of Waterloo, Canada, under C. L. Stewart. Also, the author would like to
thank J. M. Borwein for valuable comments on earlier versions of the paper.
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Habib Muzaffar: Department of Mathematics, The University of Toledo, 2801 W. Bancroft Street,
MS-942, Toledo, OH 43606-3390, USA
E-mail address: habib.muzaffar@utoledo.edu