Evaluate the series ∞ X n=1 n ∞ X k=n 1 k2 −1 − 1 2n

Problem 12134

(American Mathematical Monthly, Vol.126, October 2019) Proposed by P. Bracken (USA).

Evaluate the series

∞

X

n=1

n

∞

X

k=n

1

k² −1 − 1 2n

! .

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

Solution. By Stolz-Cesaro,

N →∞lim P^∞

k=N1/k²−1/N

1/N² lim

N →∞

−1/N²−1/(N + 1) + 1/N 1/(N + 1)²−1/N² = 1

2, and therefore

∞

X

k=N

1 k² = 1

N + 1

2N²+ o(1/N²).

Hence, for N ≥ 1, we have that

N

X

n=1

n

∞

X

k=n

1

k² −1 − 1 2n

!

=

N

X

n=1

n

N

X

k=n

1 k²+

N

X

n=1

n

∞

X

k=N +1

1

k² −N −H^N 2

=

N

X

k=1

1 k²

k

X

n=1

n +N (N + 1) 2

∞

X

k=N +1

1

k² −N −H^N 2

=

N

X

k=1

k(k + 1)

2k² +N (N + 1) 2

 1 N + 1

2N²+ o(1/N²) − 1 N²



−N −H^N 2

=N 2 +H^N

2 +N + 1

2 −N + 1

4N + o(1) − N −H^N 2

=1 4 + o(1) where H^N =P^N

k=11/k. Finally, as N → ∞, we find that the required sum is equal to 1/4.