Problem 11149
(American Mathematical Monthly, Vol.112, April 2005) Proposed by M. Ivan and I. Rasa (Romania).
Leta > 0. Find
n→∞lim n log(1 + log(1 + (· · · log(1 + a/n) · · · ))) where the parentheses are nested to depthn.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let f (z) be the principal complex logarithm log(1 + z). We denote by fn(z) the nth-iterate of f in the right half plane H = {z ∈ C : Re(z) > 0} (note that f (H) ⊂ H). We will prove that
n→∞lim nfn(z/n) = 2z
z + 2 for z ∈ H (which is the required limit when z is a real positive number).
Since |f (z)| ≤ |z| for z ∈ H (see the final remark) then
|nfn(z/n)| = n|f (fn−1(z/n))| ≤ n|fn−1(z/n)| ≤ n|z/n| = |z|
and therefore the sequence of analytic functions {nfn(z/n)} is locally bounded on H. If we show that this sequence converges to the analytic function g(z) = 2z/(z + 2) in (0, 2) (which has an accumulation point in H), then by Vitali’s theorem the sequence converges uniformly on compact sets of H to g.
LetP∞
k=1an,kzkbe the expansion of fn(z) at zero. Then a1,k= (−1)k+1/k and, since fn= fn−1◦ f and f(k)(0) = (−1)k+1(k − 1)!, by the Fa`a di Bruno’s theorem we are able find a recursion formula for the coefficients for n ≥ 2:
an,k = fn(k)(0)
k! =X
π(k)
f(1)(0)r1f(2)(0)r2
· · ·f(k)(0)rk
r1!r2! · · · rk! · (1!)r1(2!)r2· · · (k!)rk · fn−1(r) (0)
= X
π(k)
r
r1, r2, . . . , rk
· (−1)k+r
1r12r2· · · krk · an−1,r
where the sum runs over all partitions π(k) of the positive integer k: k = 1r1+ 2r2+ · · · + krk, ri
denotes the number of parts of size i, and r = r1+ r2+ · · · + rk is the total number of parts.
It is easy to determine an explicit formula for the first coefficients:
an,1 = an−1,1 = a1,1 = 1
an,2 = an−1,2− an−1,1/2 = an−1,2− 1/2 = −n/2
an,3 = an−1,3− an−1,2+ an−1,1/3 = an−1,3− (n − 1)/2 + 1/3 = n2/4 + n/12.
Now we show by induction with respect to k that
an,k =
−n 2
k−1
+ Qk(n)
where Qk is a polynomial of degree k − 2. The inductive step: by the recursion formula
an,k= an−1,k− k − 1 2
an−1,k−1+
k−2
X
r=1
ck,ran−1,r
where the first two terms are associated to the partitions 1 · k and 1 · (k − 2) + 2 · 1. Hence
an,k = a1,k+
n−1
X
j=1
− k − 1 2
aj,k−1+
k−2
X
r=1
ck,raj,r
!
= a1,k+
n−1
X
j=1
− k − 1 2
−j 2
k−2
+ Qk−1(j)
! +
k−2
X
r=1
ck,raj,r
!
= (k − 1)
−1 2
k−1 n−1
X
j=1
jk−2+ O
n−1
X
j=1
jk−3
=
−1 2
k−1
nk−1+ O(nk−2).
Finally
nfn(z/n) =
∞
X
k=1
an,k
nk−1 · zk=
∞
X
k=1
−1 2
k−1
+Qk(n) nk−1
! zk
and for z ∈ (0, 2) taking the limit as n goes to infinity we have that
n→∞lim nfn(z/n) = z
∞
X
k=1
−z 2
k−1
= z
1 − (−z/2) = 2z 2 + z.
Remark. | log(1 + z)| ≤ |z| for all z in the right half-plane H.
The function h(z) = log(1 + z)/z is analytic in a neighborhood of H then by the maximum modulus theorem it suffices to show that |h(z)| ≤ 1 on the boundary of the domain H ∩ {|z| < r} for r greater than some r0. On the imaginary axes z = iy:
| log(1 + z)|2= |(1/2) log(1 + y2) + i arctan y|2= (1/4)(log(1 + y2))2+ (arctan y)2≤ y2= |z|2. Moreover, on the semicircumference z = reiθ with θ ∈ [−π/2, π/2]:
| log(1 + z)| ≤ log |1 + z| + π/2 ≤ log(1 + r) + π/2 ≤ (1/2)(r − 1) + log 2 + π/2 ≤ r = |z|
for r ≥ r0= 2 log 2 + π − 1.
