X n=0 2n Cn = 5 +3π 2

Problem 11765

(American Mathematical Monthly, Vol.121, March 2014) Proposed by D. Beckwith (USA).

Let Cn be the nth Catalan number, given by Cn= _n+1^{1 2n}_n
. Show that (a)

∞

X

n=0

2ⁿ Cn

= 5 +3π

2 ; (b)

∞

X

n=0

3ⁿ Cn

= 22 + 8√ 3π.

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

It is known that for |z| < 2

f (z) := (arcsin(z/2))²= 1 2

∞

X

n=1

z²ⁿ n^{2 2n}_n
. Therefore

D_z(z³D_z(zD_zf )) = D_z z³D_z zD_z1 2

∞

X

n=1

z²ⁿ n^{2 2n}_n

!!

= D_z z³D_z

∞

X

n=1

z²ⁿ n ²ⁿ_n

!

= 2Dz

∞

X

n=1

z²ⁿ⁺²

2n n

= 4z

∞

X

n=1

z²ⁿ C_n. Hence

∞

X

n=0

z²ⁿ Cn

= 1 + 1

4zDz(z³Dz(zDzf )) = g(z) := 2(8 + z²)

(4 − z²)² +24z arcsin(z/2) (4 − z²)^5/2 , which implies that

(a)

∞

X

n=0

2ⁿ Cn

= g(√

2) = 5 +3π

2 ; (b)

∞

X

n=0

3ⁿ Cn

= g(√

3) = 22 + 8√ 3π.