n X k=1 (−1)k kd , Hn(−1, −1

Problem 11682

(American Mathematical Monthly, Vol.119, December 2012) Proposed by Ovidiu Furdui (Romania).

Compute

∞

X

n=0

(−1)ⁿ

∞

X

k=1

(−1)^k−1 n + k

!² .

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

Let

Hn(d) =

n

X

k=1

1

k^d, Hn(−d) =

n

X

k=1

(−1)^k

k^d , Hn(−1, −1) = X

1≤k<j≤n

(−1)^j+k kj ,

Hn(1, −1) = X

1≤k<j≤n

(−1)^j

kj , Hn(−1, 1) = X

1≤k<j≤n

(−1)^k kj . We first note that

an=

∞

X

k=1

(−1)^k−1 n + k

!2

= (−1)ⁿ

∞

X

k=n+1

(−1)^k−1 k

!2

= ln 2 −

n

X

k=1

(−1)^k−1 k

!2

.

So a_n is strictly decreasing to zero and the desired series converge. For any positive integer N ,

S2N = (ln 2)²+

2N

X

n=1

(−1)ⁿan= (ln 2)²+

2N

X

n=1

(−1)ⁿ ln 2 −

n

X

k=1

(−1)^k−1 k

!2

= (ln 2)²− 2(ln 2)

2N

X

n=1

(−1)ⁿ

n

X

k=1

(−1)^k−1

k +

2N

X

n=1

(−1)ⁿ

n

X

k=1

1 k²

+ 2

2N

X

n=1

(−1)ⁿ

n

X

k=1 n

X

j=k+1

(−1)^k+j kj . Moreover

2N

X

n=1

(−1)ⁿ

n

X

k=1

(−1)^k−1

k =

2N

X

k=1

(−1)^k−1 k

2N

X

n=k

(−1)ⁿ= −1 2H_N(1),

2N

X

n=1

(−1)ⁿ

n

X

k=1

1 k² =

2N

X

k=1

1 k²

2N

X

k=n

(−1)ⁿ =1 4HN(2)

2N

X

n=1

(−1)ⁿ

n

X

k=1 n

X

j=k+1

(−1)^k+j kj =

2N

X

k=1 2N

X

n=k n

X

j=k+1

(−1)^n+k+j

kj =

2N

X

k=1 2N

X

j=k+1

(−1)^k+j kj

2N

X

n=j

(−1)ⁿ

= 1

2H2N(−1, 1) +1

2H2N(−1, −1).

Hence,

S_2N = (ln 2)²+ (ln 2)H_N(1) +1

4H_2N(2) + H_2N(−1, 1) + H_2N(−1, −1)

= (ln 2)²+ (ln 2)HN(1) +1

4H2N(2) + H2N(−1)H2N(1) − H2N(1, −1) − H2N(−2) +1

2(H2N(−1))²−1

2H2N(2).

Finally

∞

X

n=0

(−1)ⁿ

∞

X

k=1

(−1)^k−1 n + k

!²

= lim

N →+∞S2N = π² 24 because

Hn(1) = ln(n) + γ + O(1/n), Hn(−1) = − ln 2 + O(1/n),

H_n(2) = π²/6 + o(1), H_n(−2) = −π²/12 + o(1), H_n(1, −1) = (ln 2)²/2 + o(1).