ψ(n − 1/φ) n2+ n − 1 = π2 2√ 5+π2tan2(√ 5π/2) √5 +4π tan(√ 5π/2) 5

Problem 11842

(American Mathematical Monthly, Vol.122, May 2015) Proposed by I. Mez˝o (China).

Let ψ be the Digamma function, that is, ψ(x) = (log Γ(x))⁰. Let φ = (1 +√

5)/2. Prove that

∞

X

n=1

ψ(n + φ) − ψ(n − 1/φ) n²+ n − 1 = π²

2√

5+π²tan²(√ 5π/2)

√5 +4π tan(√ 5π/2)

5 .

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

Since ψ(n + z) = ψ(a) +Pn−1 k=0

1

k+z, it follows that if a and b are not non-positive integers then for all integers N ≥ 0,

N

X

n=0

ψ(n + a) − ψ(n + b)

(n + a)(n + b) =(ψ(a) − ψ(b))

N

X

n=0

1 (n + a)(n + b)

+(b − a) 2





N

X

n=0

1 (n + a)(n + b)

!²

−

N

X

n=0

1

((n + a)(n + b))²



. In our case a = φ and b = −1/φ, and as N goes to infinity we obtain

∞

X

n=1

ψ(n + φ) − ψ(n − 1/φ)

n²+ n − 1 =(ψ(φ) − ψ(−1/φ)) 1 +

∞

X

n=0

1 n²+ n − 1

!

−

√5 2





∞

X

n=0

1 n²+ n − 1

!²

−

∞

X

n=0

1 (n²+ n − 1)²



. The required formula is proved as soon as we compute ψ(φ) − ψ(−1/φ) and the two series.

i) By the reflection formula,

ψ(φ) − ψ(−1/φ) = ψ(φ) − ψ(1 − φ) = − π

tan(πφ) = π tan(√ 5π/2).

ii) By the residue formula, we have that

∞

X

n=0

1

n²+ n − 1= 1 2

∞

X

n=−∞

1

n²+ n − 1= π 2

X

n∈Z

Res

 cot(πz) z²+ z − 1, z



= −π 2

 Res

 cot(πz) z²+ z − 1, −φ

 + Res

 cot(πz) z²+ z − 1, 1/φ



= −π cot(πφ)

√5 =π tan(√ 5π/2)

√5 .

ii) By the residue formula, we have that

∞

X

n=0

1

(n²+ n − 1)² =1 2

∞

X

n=−∞

1

(n²+ n − 1)² = π 2

X

n∈Z

Res

 cot(πz) (z²+ z − 1)², z



= −π 2

 Res

 cot(πz) (z²+ z − 1)², −φ

 + Res

 cot(πz) (z²+ z − 1)², 1/φ



= −π 5



− π

sin²(πφ)−2 cot(πφ)

√5



=π²

5 +π²tan²(√ 5π/2)

5 −2π tan(√ 5π/2) 5√

5 .