Problem 12051
(American Mathematical Monthly, Vol.125, June-July 2018) Proposed by P. Ribeiro (Portugal).
Prove
∞
X
n=0
2n n
1
4n(2n + 1)3 = π3
48 +π ln2(2)
4 .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We have that Z 1
0
x2nln(x) dx =
x2n+1
ln(x)
(2n + 1)− 1 (2n + 1)2
1
0
= − 1
(2n + 1)2 and, for x ∈ [0, 1],
∞
X
n=0
2n n
x2n
4n(2n + 1) = arcsin(x)
x .
Hence
∞
X
n=0
2n n
1
4n(2n + 1)3 = − Z 1
0
arcsin(x) ln(x)
x dx
= −
arcsin(x)ln2(x) 2
1
0
+1 2
Z 1
0
ln2(x)
√1 − x2dx
= 1 2
Z 1 0
ln2(x)
√1 − x2dx = 1 2
Z π/2 0
ln2(sin(t)) dt = π3
48 +π ln2(2) 4 where the last integral is known.
For the sake of completeness, we give here a short proof. Since lnn(1 + z) is holomorphic in the unit disc and it is zero for z = 0, it follows that
0 = Z π
−π
lnn(1 + eiθ) dθ = Z π
−π
(ln(2 cos(θ/2) + iθ/2)ndθ = 2 Z π/2
−π/2
(ln(2 cos(t)) + it)ndt.
By taking the real part we get
⌊n/2⌋
X
k=0
(−1)k n 2k
Z π/2 0
t2klnn−2k(2 cos t) dt = 0.
For n = 1 and for n = 2, we get Z π/2
0
ln(cos t) dt = −π
2ln(2) ,
Z π/2 0
ln2(2 cos t) dt = Z π/2
0
t2dt =π3 24. Hence
Z π/2 0
ln2(sin(t)) dt = Z π/2
0
ln2(sin(π/2 − t)) dt = Z π/2
0
ln2(cos(t)) dt
= Z π/2
0
ln2(2 cos(t)) dt − 2 ln(2) Z π/2
0
ln(cos(t)) dt −π ln2(2) 2
= π3
24+ π ln2(2) −π ln2(2) 2 = π3
24 +π ln2(2)
2 .
