Problem 11768
(American Mathematical Monthly, Vol.121, April 2014) Proposed by Ovidiu Furdui (Romania).
Let f be a bounded continuous function mapping [0, +∞) to itself. Find
n→∞lim n
n
s Z +∞
0
fn+1(x)e−xdx − n s
Z +∞
0
fn(x)e−xdx
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let h(t) = f (− ln(t)) then h is a non-negative bounded continuous function in (0, 1].
Let an =R1
0 hn(t)dt then
L := lim
n→∞n
n
s Z +∞
0
fn+1(x)e−xdx − n s
Z +∞
0
fn(x)e−xdx
= lim
n→∞n √n
an+1− √n an .
We will show that L = M ln(M ) where M := sup
x∈[0,+∞)
f (x) = sup
t∈(0,1]
h(t) ≥ 0 (if M = 0, then L = 0).
If M = 0 then f and h are identically zero and the limit is trivial. Let us assume that M > 0.
For 0 < ε < M , there is a non-empty interval I in (0, 1] such that for all t ∈ I, h(t) ≥ M − ε > 0.
Hence
(M − ε)|I|1/n= ((M − ε)n|I|)1/n≤ √n
an≤ (Mn|(0, 1]|)1/n = M, and, |I|1/n→ 1 (|I| > 0) implies that √n
an→ M .
Now we consider the sequence an+1/an. It is bounded because an+1
an = 1 an
Z 1 0
hn+1(t)dt ≤ M an
Z 1 0
hn(t)dt = M.
It is increasing because, by Cauchy-Schwarz inequality,
a2n+1=
Z 1 0
hn+22 (t)hn2(t)dt
2
≤ Z 1
0
hn+2(t)dt Z 1
0
hn(t)dt = an+2an.
So an+1/an has a limit M0. By Stolz-Cesaro theorem, we find that M0= M ,
ln(M ) = lim
n→∞ln(√n
an) = lim
n→∞
ln(an) n = lim
n→∞(ln(an+1) − ln(an)) = lim
n→∞ln an+1
an
= ln(M0).
Finally,
n→∞lim n √n
an+1− √n
an = lim
n→∞n√n an
exp 1
nln an+1
an
− 1
= lim
n→∞
√n
anln an+1
an
= M ln(M ).
