On Sums Involving Binomial Coefficients

S. Amghibech Rue Wilfrid-Légaré Québec, QC Canada

                     amghibech@hotmail.com

Abstract We give closed forms for the series ∑_{m=1}^{∞} \frac{(2x)^{2m+2k}}{m^{2}(m+k)\binom{2m}{m}} and ∑_{m=1}^{∞} \frac{(2x)^{2m}(−1)^{m+k}}{m^{2}(m+k)\binom{2m}{m}} for

integers k ≥ 0.

1 Introduction

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

∑_{m≥1} \frac{(2x)^{2m}}{m\binom{2m}{m}} = \frac{2x}{\sqrt{1 − x^{2}}} arcsin x,  (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

∑_{m=1}^{∞} \frac{(2t)^{2m+2k}}{m(2m + 2k)\binom{2m}{m}} = arcsin(t)^{2}\binom{2k}{k} + ∑_{j=1}^{k} \binom{2k}{k − j}\frac{(−1)^{j+1}}{j^{2}}
+ ∑_{j=1}^{k}(−1)^{j}\binom{2k}{k − j} (\frac{2 arcsin t sin(2j arcsin t)}{j} + \frac{cos(2j arcsin t)}{j^{2}}) .

(b) If we replace t by \sqrt{−1}t, we get another form

∑_{m=1}^{∞} \frac{(−1)^{m+k}(2t)^{2m+2k}}{m(2m + 2k)\binom{2m}{m}} = − log(t + \sqrt{1 + t^{2}})^{2}\binom{2k}{k} + ∑_{j=1}^{k} \binom{2k}{k − j}\frac{(−1)^{j+1}}{j^{2}}

+ ∑_{j=1}^{k}(−1)^{j}\binom{2k}{k − j}  (\frac{(−t + \sqrt{t^{2} + 1})^{2j} − (t + \sqrt{t^{2} + 1})^{2j}}{j} ) log(t + \sqrt{t^{2} + 1})

+ ∑_{j=1}^{k}(−1)^{j}\binom{2k}{k − j}  (\frac{(−t + \sqrt{t^{2} + 1})^{2j} + (t + \sqrt{t^{2} + 1})^{2j})}{2j^{2}}) .

Theorem 2.2. (a) For t ∈ (−1, 1) we have

∑_{m=1}^{∞} \frac{(2t)^{2m+2k}}{m^{2}(2m + 2k)\binom{2m}{m}} = \frac{−(arcsin t)^{2}}{k} \binom{2k}{k} + \frac{(2t)^{2k}}{k} (arcsin t)^{2} − ∑_{j=1}^{k} \binom{2k}{k − j}\frac{(−1)^{j+1}}{kj^{2}}

− ∑_{j=1}^{k}(−1)^{j}\binom{2k}{k − j} (\frac{2 arcsin t sin(2j arcsin t)}{kj} + \frac{cos(2j arcsin t)}{kj^{2}}) .

(b) The identity in (a) is valid if we replace t by it and arcsin(it) by i log(1 + \sqrt{1 + t^{2}}):

∑_{m=1}^{∞} \frac{(2t)^{2m+2k}(−1)^{m+k}}{m^{2}(2m + 2k)\binom{2m}{m}} = (\frac{1}{k}\binom{2k}{k} − \frac{(2t)^{2k}}{k} ) (log(1 + \sqrt{1 + t^{2}}))^{2} − ∑_{j=1}^{k} \binom{2k}{k − j}\frac{(−1)^{j+1}}{kj^{2}}

− ∑_{j=1}^{k}(−1)^{j}\binom{2k}{k − j}  (\frac{(−t + \sqrt{t^{2} + 1})^{2j} − (t + \sqrt{t^{2} + 1})^{2j}}{jk} ) log(t + \sqrt{t^{2} + 1})

− ∑_{j=1}^{k}(−1)^{j}\binom{2k}{k − j}  (\frac{(−t + \sqrt{t^{2} + 1})^{2j} + (t + \sqrt{t^{2} + 1})^{2j})}{2kj^{2}}) .

We use the basic method to prove various combinatorial identities where binomial coefficients occur in the denominator. As a particular consequence of our results we get the sums S_{1}(k), S_{2}(k), T_{1}(k), T_{2}(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, then

∑_{m≥1} \frac{(2x)^{2m}}{m\binom{2m}{m}} = \frac{2x}{\sqrt{1 − x^{2}}} arcsin x.  (2)

If we replace x by \sqrt{−1}x, we get another form of the identity (2):

∑_{m≥1} \frac{(−1)^{m}(2x)^{2m}}{m\binom{2m}{m}} = \frac{−2x}{\sqrt{1 + x^{2}}} log(x + \sqrt{1 + x^{2}})

If we multiply both members of the equation (2) by x^{2k−1} and then integrate, we obtain

∑_{m≥1} \frac{2^{2m}t^{2m+2k}}{m(2m + 2k)\binom{2m}{m}} = ∫_{0}^{t} \frac{2x^{2k}}{\sqrt{1 − x^{2}}} arcsin xdx.

The left-hand side of this equation can be written

∫_{0}^{arcsin t} 2x(sin(x))^{2k}dx = \frac{(−1)^{k}}{2^{2k−1}} ∑_{j=0}^{2k} \binom{2k}{j}(−1)^{j} ∫_{0}^{arcsin t} x exp(2ix(k − j))dx

from which we get

∫_{0}^{arcsin t} 2x(sin(x))^{2k}dx = \frac{(arcsin t)^{2}}{2^{2k}} \binom{2k}{k}
+ \frac{1}{2^{2k−1}} ∑_{j=1}^{k}(−1)^{j}\binom{2k}{k − j} (\frac{τ sin(2jτ)}{j} + \frac{cos(2jτ)}{2j^{2}} − \frac{1}{2j^{2}} )

where τ = arcsin(t). Finally we get

∑_{m≥1} \frac{(2t)^{2m+2k}}{m(2m + 2k)\binom{2m}{m}} = (arcsin t)^{2}\binom{2k}{k} + ∑_{j=1}^{k} \binom{2k}{k − j}\frac{(−1)^{j+1}}{j^{2}}
+ ∑_{j=1}^{k}(−1)^{j}\binom{2k}{k − j} (\frac{2(arcsin t) sin(2j arcsin t)}{j} + \frac{cos(2j arcsin t)}{j^{2}}) ,

which gives Theorem 2.1. Proof of Theorem 2.1.

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

∑_{m≥1} \frac{(2x)^{2m}}{m^{2}\binom{2m}{m}} = 2(arcsin x)^{2}

If we multiply both members of this equation by x^{2k−1} and then integrate, we obtain

∑_{m≥1} \frac{2^{2m}t^{2m+2k}}{m^{2}(2m + 2k)\binom{2m}{m}} = ∫_{0}^{t} 2x^{2k−1}(arcsin x)^{2}dx.

The left-hand side of can be written

∫_{0}^{t} 2x^{2k−1}(arcsin x)^{2}dx = \frac{t^{2k}}{k} (arcsin t)^{2} − \frac{1}{k} ∫_{0}^{t} \frac{2x^{2k}}{\sqrt{1 − x^{2}}} arcsin xdx.

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

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 in Journal of Integer Sequences, January 15 2007.

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