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23 11

Article 07.2.1

Journal of Integer Sequences, Vol. 10 (2007), 2

3 6 1 47

On Sums Involving Binomial Coefficients

S. Amghibech

Rue Wilfrid-L´egar´e

Qu´ebec, QC

Canada

[email protected]

Abstract

We give closed forms for the series P∞ m=1

(2x)2m+2k

m2(m+k)(2m m)

and P∞ m=1

(2x)2m

(−1)m+k

m2(m+k)(2m m)

for integersk0.

1

Introduction

D. H. Lehmer [1] studied various series with binomial coefficients in the denominator, for example,

X

m≥1

(2x)2m m¡2mm¢ =

2x √

1x2 arcsinx, (1)

valid for |x|<1. In this note we consider some related results.

2

Main Results

The main results of the paper can be stated as follows:

Theorem 2.1. (a) For t(1,1) we have

X

m=1

(2t)2m+2k m(2m+ 2k)¡2m

m

¢ = arcsin(t)

2

µ 2k

k

¶ +

k

X

j=1

µ 2k kj

(1)j+1 j2

+

k

X

j=1

(1)j µ

2k kj

¶ µ

2 arcsintsin(2jarcsint)

j +

cos(2jarcsint)

j2

.

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(b) If we replace t by1t, we get another form

We use the basic method to prove various combinatorial identities where binomial coeffi-cients occur in the denominator. As a particular consequence of our results we get the sums

S1(k), S2(k), T1(k), T2(k) computed in [2] in a compact form.

3

Proofs of the Theorems

Proof of Theorem 2.1.

We start with the following identity, which was discovered by D. H. Lehmer [1]. If|x|<1,

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If we replace x by√1x, we get another form of the identity (2):

X

m≥1

(1)m(2x)2m m¡2m

m

¢ =

−2x √

1 +x2log(x+

1 +x2)

If we multiply both members of the equation (2) by x2k−1 and then integrate, we obtain

X

m≥1

22mt2m+2k m(2m+ 2k)¡2m

m

¢ =

Z t

0

2x2k √

1x2 arcsinxdx.

The left-hand side of this equation can be written

Z arcsint

0

2x(sin(x))2kdx= (−1)k

22k−1 2k

X

j=0

µ 2k

j

¶ (1)j

Z arcsint

0

xexp(2ix(kj))dx

from which we get

Z arcsint

0

2x(sin(x))2kdx = (arcsint)2

22k

µ 2k

k

+ 1

22k−1 k

X

j=1

(1)j

µ 2k kj

¶ µ

τsin(2jτ)

j +

cos(2jτ)

2j2 −

1 2j2

where τ = arcsin(t). Finally we get

X

m≥1

(2t)2m+2k m(2m+ 2k)¡2m

m

¢ = (arcsint)

2

µ 2k

k

¶ +

k

X

j=1

µ 2k kj

(1)j+1 j2

+

k

X

j=1

(1)j

µ 2k kj

¶ µ

2(arcsint) sin(2jarcsint)

j +

cos(2jarcsint)

j2

,

which gives Theorem2.1.

Proof of Theorem 2.1.

If we integrate (2) from 0 to x we obtain

X

m≥1

(2x)2m m2¡2m

m

¢ = 2(arcsinx)

2

If we multiply both members of this equation byx2k−1 and then integrate, we obtain

X

m≥1

22mt2m+2k m2(2m+ 2k)¡2m

m

¢ =

Z t

0

2x2k−1(arcsinx)2dx.

The left-hand side of can be written Z t

0

2x2k−1(arcsinx)2dx= t

2k

k (arcsint) 2

− 1k

Z t

0

2x2k √

1x2 arcsinxdx.

By using the proof of Theorem 2.1 we get Theorem 2.2.

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References

[1] D. H. Lehmer. Interesting series involving the central binomial coefficient. Amer. Math.

Monthly 92 (1985), 449–457.

[2] Jin-Hua Yang and Feng-Zhen Zhao, Sums involving binomial coefficients, Journal of

Integer Sequences 9 (2006), Article 06.4.2.

2000 Mathematics Subject Classification: Primary 11B65.

Keywords: binomial coefficients, integral.

Received December 26 2006; revised version received January 11 2007. Published inJournal

of Integer Sequences, January 15 2007.

Return to Journal of Integer Sequences home page.

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