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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

(2)

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 = re 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.

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