Problem 12094
(American Mathematical Monthly, Vol.126, February 2019) Proposed by P. F. Refolio (Spain).
Prove
∞
X
n=0 2n
n
2
16n(n + 1)3 = 16 log(2) − 32G π +48
π − 16 whereG =P∞
k=0 (−1)k
(2k+1)2 is the Catalan’s constant.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let
Cn :=
2n n
n + 1 = 22n+1
π B(n + 1/2, 3/2) = 2 · 4n π
Z 1 0
xnr 1 − x x dx be the nth Catalan number where B(s, t) denotes the Beta function. For x ∈ (0, 1),
∞
X
n=0
Cn(x4)n n + 1 = 1
x Z x
0
∞
X
n=0
Cn
t 4
n
dt = 2 x
Z x 0
1 −√ 1 − t t dt = 4
x
1 −√
1 − x + log 1 +√ 1 − x 2
. Therefore, by the above integral representation of Cn, after letting x = cos2(t) we find
S : = X∞ n=0
2n n
2 16n(n + 1)3 =
Z 1 0
r 1 − x x
X∞ n=0
Cn(x4)n n + 1 dx
= 8 π
Z 1 0
√1 − x − 1 + x +√
1 − x log(1+√21−x)
x3/2 dx
= 16 π
Z π/2 0
tan2(t) − tan2(t) sin(t) + tan2(t) log 1 + sin(t) 2
dt.
Now we have that Z π/2
0
tan2(t) − tan2(t) sin(t) dt = tan(t)(1 − sin3(t)) − (2 + sin2(t)) cos(t) − tπ/2
0 = 2 −π 2, and
Z π/2 0
(1 + tan2(t)) log 1 + sin(t) 2
dt =
Z π/2 0
D(tan(t)) log 1 + sin(t) 2
dt
= 0 − Z π/2
0
sin(t)
1 + sin(t)dt = −
t + 2
1 + tan(t/2)
π/2 0
= 1 −π 2. Moreover, since
log(1 + cos(t)) = log(1 + eit) + log(1 + e−it) − log(2)
=
∞
X
k=1
(−1)k−1eikx
k +
∞
X
k=1
(−1)k−1e−ikx
k − log(2) = 2
∞
X
k=1
(−1)k−1cos(kx)
k − log(2)
it follows that Z π/2
0
log 1 + sin(t) 2
dt =
Z π/2
0 log(1 + cos(t)) dt −π log(2)
2 = 2
∞
X
k=1
(−1)k−1 k
Z π/2
0 cos(kx) dx − π log(2)
= 2
∞
X
k=1
(−1)k−1sin(kπ/2)
k2 − π log(2) = 2G − π log(2).
Finally
S = 16 π
2 − π 2
+ 1 − π
2
− (2G − π log(2))
= 16 log(2) − 32G π +48
π − 16.
