1 n n X k=1 cot(θk/2) sin θk (x − cos θk)2+ sin2θk where wk= exp(iθk) with θk = (2k − 1)π/n for k = 1

Problem 11873

(American Mathematical Monthly, Vol.122, December 2015) Proposed by E. J. Ionascu (USA).

Show that forn ∈ N with n ≥ 2,

n

X

k=1



1 − 2k − 1 n



cot(2k − 1)π

2n =

n−1

X

k=1

csckπ n .

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

First solution. We will show the identity by computing the integral

In:=

Z 1 0

1 − xⁿ⁻¹ (1 − x)(1 + xⁿ)dx in two different ways.

By using the partial fraction decomposition, for x ≥ 0,

1 − xⁿ⁻¹ (1 − x)(1 + xⁿ) =

n

X

k=1

1−wⁿ_k⁻¹ (1−wk)nwⁿ_k⁻¹

x − wk = −i n

n

X

k=1

cot(θk/2) x − wk

= 1 n

n

X

k=1

cot(θk/2) · Im

 1

x − wk



= 1 n

n

X

k=1

cot(θk/2) sin θk

(x − cos θk)²+ sin²θk

where wk= exp(iθk) with θk = (2k − 1)π/n for k = 1, . . . , n are the n-th roots of −1. Hence

In = 1 n

n

X

k=1

cot(θk/2) Z 1

0

sin θk

(x − cos θk)²+ sin²θk

dx

= 1 n

n

X

k=1

cot(θk/2)



arctan x − cos θk

sin θk

1 0

= π 2n

n

X

k=1



1 − 2k − 1 n



cot(2k − 1)π 2n . On the other hand, we have that,

In = Z 1

0

Pn−1 k=1x^k−1 1 + xⁿ dx = 1

2

n−1

X

k=1

Z 1 0

x^k−1 1 + xⁿdx +

Z 1 0

x^n−k−1 1 + xⁿ dx



=1 2

n−1

X

k=1

Z 1 0

x^k−1 1 + xⁿ dx +

Z 1 +∞

(1/x)^n−k−1 1 + (1/x)ⁿ d(1/x)



=1 2

n−1

X

k=1

Z 1 0

x^k−1 1 + xⁿ dx +

Z +∞

1

x^k−1 1 + xⁿ



=1 2

n−1

X

k=1

Z +∞

0

x^k−1

1 + xⁿ dx = π 2n

n−1

X

k=1

csckπ n .

The proof is complete as soon as we compare the two formulas for In. 

Second solution. It is known that for z = e^iθ with θ ∈ R,

cot θ = i z + z⁻¹

z − z⁻¹ = i + 2i

z²−1 and csc θ = 2i

z − z⁻¹ = 2iz z²−1 Hence, by letting z = e^iπ/n, the identity is equivalent to

n

X

k=1



1 − 2k − 1 n

 

i + 2i z^2k−1−1



=

n−1

X

k=1

2iz^k z^2k−1 that is

n

X

k=1



1 −2k − 1 n



· 1

z^2k−1−1 =

n−1

X

k=1

z^k z^2k−1. The above identity holds because

n

X

k=1



1 −2k − 1 n



· 1

z^2k−1−1 =

2n−1

X

k=1

 1 − k

n



· 1 z^k−1 −

n−1

X

k=1

 1 − 2k

n



· 1 z^2k−1

=

2n−1

X

k=1

1 z^k−1 −1

n

2n−1

X

k=1

k z^k−1 −1

2

n−1

X

k=1

1 z^k−1 +1

2

n−1

X

k=1

1 z^k+ 1

+1 n

n−1

X

k=1

k z^k−1 −1

n

n−1

X

k=1

k z^k+ 1

=

2n−1

X

k=n

1 z^k−1 −1

n

2n−1

X

k=n

k z^k−1 +1

2

n−1

X

k=1

1 z^k−1 +1

2

n−1

X

k=1

1 z^k+ 1

−1 n

2n−1

X

k=n

k − n z^k−n+ 1

zⁿ=−1

=

2n−1

X

k=n

1 z^k−1 −1

n

2n−1

X

k=n

k z^k−1 +

n−1

X

k=1

z^k z^2k−1+ 1

n

2n−1

X

k=n

k − n z^k−1

=

n−1

X

k=1

z^k z^2k−1.