Problem 11222
(American Mathematical Monthly, Vol.113, May 2006) Proposed by J. Sondow (USA).
Fix an integerB ≥ 2, and let s(n) denote the sum of the base-B digits of n. Prove that
∞
Y
n=0
Y
k odd 0 < k < B
nB + k nB + k + 1
(−1)s(n)
= 1
√B.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let
P := Y
n≥0
Y
kodd 0 < k < B
nB + k nB + k + 1
ǫ(n)
and Q := Y
n≥1
Y
keven 0 ≤ k < B
nB + k nB + k + 1
ǫ(n)
where ǫ(n) := (−1)s(n). It suffices to prove that P2= 1/B.
Then
P · Q = Y
kodd 0 < k < B
k k + 1
ǫ(0)
·Y
n≥1
nB nB + B
ǫ(n)
= Y
kodd 0 < k < B
k k + 1
+1
·Y
n≥1
n n + 1
ǫ(n)
.
If n = mB + k with 0 ≤ k < B then s(n) = s(B) + k and ǫ(n) = (−1)k· ǫ(m)
Y
n≥1
n n + 1
ǫ(n)
= Y
0<k<B
k k + 1
ǫ(k)
· Y
m≥1
Y
0≤k<B
mB + k mB + k + 1
ǫ(mB+k)
= Y
kodd 0 < k < B
k k + 1
−1
· Y
keven 0 < k < B
k k + 1
+1
·
Y
m≥1
Y
kodd 0 < k < B
mB + k mB + k + 1
−ǫ(m)
· Y
m≥1
Y
keven 0 ≤ k < B
mB + k mB + k + 1
ǫ(m)
= Y
kodd 0 < k < B
k k + 1
−1
· Y
keven 0 < k < B
k k + 1
+1
· Y
kodd 0 < k < B
k k + 1
+1
· P−1· Q
= Y
keven 0 < k < B
k k + 1
+1
· P−1· Q
Therefore
P ·Q = Y
kodd 0 < k < B
k k + 1
+1
· Y
keven 0 < k < B
k k + 1
+1
·P−1·Q = Y
0<k<B
k k + 1
!
·P−1·Q = 1
B·P−1·Q
that is P2= 1/B. Note that all the infinite products are convergent by Abel’s theorem.
