(1)Problem 11810 (American Mathematical Monthly, Vol.122, January 2015) Proposed by O

Problem 11810

(American Mathematical Monthly, Vol.122, January 2015) Proposed by O. Furdui (Romania).

Find ∞

X

n=1

H_n ζ(3) −

n

X

k=1

1 k³

!

where H_n =Pn k=11/k.

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

We have that

∞

X

n=1

H_n ζ(3) −

n

X

k=1

1 k³

!

=

∞

X

n=1

H_n

∞

X

k=n+1

1 k³ =

∞

X

k=2

1 k³

k−1

X

n=1

H_n=

∞

X

k=2

1 k³

k−1

X

n=1 n

X

j=1

1 j

=

∞

X

k=2

1 k³

k−1

X

j=1

1 j

k−1

X

n=j

1 =

∞

X

k=2

1 k³

k−1

X

j=1

k − j j

=

∞

X

k=2

1 k²

k−1

X

j=1

1 j −

∞

X

k=2

k − 1 k³

= ζ(2, 1) − ζ(2) + ζ(3) = 2ζ(3) − ζ(2)

where in the last step we used the famous identity of Euler ζ(2, 1) = ζ(3).