X m=1 X a>m X b>m 1 a2b2 − S

Problem 11885

(American Mathematical Monthly, Vol.123, January 2016) Proposed by C. I. V˘alean (Romania).

Prove that

∞

X

p=1

∞

X

n=1

∞

X

m=1

1

(m + n)⁴+ ((m + n)(m + p))² = 3

2ζ(3) −5 4ζ(4).

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

Let S be the left-hand side, then by setting a = m + n and b = m + p, we have that by symmetry,

S=

∞

X

m=1

X

a>m

X

b>m

1

a²(a²+ b²) =

∞

X

m=1

X

a>m

X

b>m

 1 a²b²−

1 b²(a²+ b²)



=

∞

X

m=1

X

a>m

X

b>m

1 a²b² − S.

Hence,

S= 1 2

∞

X

m=1

X

a>m

X

b>m

1 a²b² =

∞

X

m=1

X

a>b>m

1 a²b² +1

2

∞

X

m=1

X

a>m

1 a⁴

= X

a>b≥1

1 a²b²

b−1

X

m=1

1 +1 2

X

a≥1

1 a⁴

a−1

X

m=1

1

= X

a>b≥1

1 a²b −

X

a>b≥1

1 a²b² +1

2 X

a≥1

1 a³−

1 2

X

a≥1

1 a⁴

= ζ(2, 1) − ζ(2, 2) +1

2ζ(3) −1

2ζ(4) = 3

2ζ(3) − 5 4ζ(4), because the following Multiple Zeta Values are known,

ζ(2, 1) = ζ(3) and ζ(2, 2) = 3 4ζ(4).