Show that forn ≥ 1, n X k=1 6n + 1 6k − 2  B6k−2= −6n + 1 6 , whereBn denotes thenth Bernoulli number

Problem 11791

(American Mathematical Monthly, Vol.121, August-September 2014) Proposed by M. ˘Stofka (Slovakia).

Show that forn ≥ 1,

n

X

k=1

6n + 1 6k − 2



B6k−2= −6n + 1 6 , whereBn denotes thenth Bernoulli number.

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

The generating function of the Bernoulli numbers is f (z) = z

e^z−1 =

∞

X

n=0

Bnzⁿ n! . Let ω = e^2πi/3, and let F (z) := f (z) + ω²f (ωz) + ωf (ω²z). Then

F (z) =

∞

X

n=0

Bnzⁿ

n! 1 + ωⁿ⁺²+ ω²ⁿ⁺¹
= 3

∞

X

n=0

B3n+1z³ⁿ⁺¹ (3n + 1)! = −3z

2 + 3

∞

X

n=1

B6n−2z⁶ⁿ⁻² (6n − 2)! , where in the last step we used the fact that B1= −1/2 and Bn = 0 for any odd integer n > 1.

On the other hand, letting

g(z) := e^z+ e^ωz+ e^ω2z= 3

∞

X

n=0

z³ⁿ (3n)! = 3

∞

X

n=0

z⁶ⁿ (6n)! + 3

∞

X

n=0

z⁶ⁿ⁺³ (6n + 3)!, we get

F (z) = z

 1

e^z−1+ 1

e^ω2z−1+ 1 e^ωz−1



=z(3 + e^−z−2e^z+ e^−ωz−2e^ωz+ e^−ω2z−2e^ω2z) e^z−e^−z+ e^ωz−e^−ωz+ e^ω2z−e^−ω2z

=z(3 − 2g(z) + g(−z)) g(z) − g(−z) =

−3

∞

X

n=1

z⁶ⁿ⁺¹ (6n)! −9

∞

X

n=0

z⁶ⁿ⁺⁴ (6n + 3)!

6

∞

X

n=0

z⁶ⁿ⁺³ (6n + 3)!

.

Therefore, we obtain the following identity 6

∞

X

n=0

z⁶ⁿ⁺³

(6n + 3)!· −3z 2 + 3

∞

X

n=1

B6n−2z⁶ⁿ⁻² (6n − 2)!

!

= −3

∞

X

n=1

z⁶ⁿ⁺¹ (6n)! −9

∞

X

n=0

z⁶ⁿ⁺⁴ (6n + 3)!, that is

∞

X

n=0

z⁶ⁿ⁺³ (6n + 3)!·

∞

X

n=1

B6n−2z⁶ⁿ⁻² (6n − 2)! = −1

6

∞

X

n=1

(6n + 1)z⁶ⁿ⁺¹ (6n + 1)! , and after extracting the coefficient of z⁶ⁿ⁺¹/(6n + 1)!, we finally have

n

X

k=1

6n + 1 6k − 2



B6k−2= −6n + 1 6 .