Problem 12180
(American Mathematical Monthly, Vol.127, April 2020) Proposed by P. F. Refolio (Spain).
Prove
X∞ n=0
4n 2n
2
28n(2n + 1) = 2 π −
√2 C2
π3/2 + π1/2
√2 C2,
whereC =R∞
0 t−1/4e−t= Γ(3/4).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Since
Z π/2 0
sin2n(x) dx = π 22n+1
2n n
and, for |z| < 1,
∞
X
n=0 4n 2nz2n 42n(2n + 1) = 1
z Z z
0
f (z) + f (−z)
2 dz =
√1 + z −√ 1 − z
z =
√2 p1 +√
1 − z2 with f (z) =P∞
n=0 2n
n
zn 4n =√1
1−z, it follows that
∞
X
n=0 4n 2n
2
28n(2n + 1) = 2 π
∞
X
n=0 4n 2n
42n(2n + 1)
Z π/2 0
sin4n(x) dx
= 2√ 2 π
Z π/2 0
dx r
1 + q
1 − sin4(x)
[sin2(x) = 2pt(1 − t)]
= 1
√2π Z 1/2
0
t1/2+ (1 − t)1/2 t3/4(1 − t)5/4 dt
= 1
√2π Z 1/2
0
t−1/4(1 − t)−5/4dt + Z 1/2
0
t−3/4(1 − t)−3/4dt
!
= 1
√2π
2√
2 − 2B1/2(3/4, 3/4) + B1/2(1/4, 1/4) where
Bx(a, b) = Z x
0
ta−1(1 − t)b−1dt is the so-called Incomplete Beta Function and we used the fact that
d dt
4t3/4(1 − t)−1/4
= t−1/4(1 − t)−5/4+ 2t−1/4(1 − t)−1/4. By noting that
B1/2(a, b) = B1(a, b)
2 = Γ(a)Γ(b) 2Γ(a + b),
Γ(3/2) = Γ(1/2)/2 and, by the Euler’s reflection formula Γ(z)Γ(1 − z) =sin(πz)z , Γ(1/2) =√
π and Γ(1/4)Γ(3/4) =√ 2π, we finally find
X∞ n=0
4n 2n
2
28n(2n + 1)= 1
√2π
2√
2 − Γ(3/4)2
Γ(3/2) +Γ(1/4)2 2Γ(1/2)
= 2 π−
√2 Γ(3/4)2
π3/2 + π1/2
√2 Γ(3/4)2.
