Problem 12000
(American Mathematical Monthly, Vol.124, October 2017) Proposed by M. Sawhney (USA).
LetHk =Pk
i=11/i. Prove that the function f : R → R defined by f (x) = 1 +
∞
X
n=1
xn Qn
k=1Hk has no real zeroes.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Note that the radius of convergence of the power series f (x) = 1 +P∞
n=1anxn is R = limn→∞
an
an+1
= limn→∞Hn+1= +∞.
Therefore f is well-defined and continuous in R. Assume that f has at least a real zero and let t := sup{x ∈ R : f (x) = 0}.
Note that t < 0 because f (x) ≥ 1 for x ≥ 0. Moreover, by continuity, f (t) = 0 and f (x) > 0 for x ∈ (t, 0]. Hence
I :=
Z 0
t
f (x) − f (t) x − t dx =
Z 0
t
f (x)
x − tdx > 0.
On the other hand, since power series can be integrated term-by-term on an interval lying inside the interval of convergence, we have that
I = Z 0
t
∞
X
n=1
xn− tn (x − t)Qn
k=1Hk dx
=
∞
X
n=1
1 Qn
k=1Hk Z 0
t
xn− tn
x − t dx (s := x/t)
= −
∞
X
n=1
tn Qn
k=1Hk Z 1
0
sn− 1 s − 1 ds
= −
∞
X
n=1
tn Qn
k=1Hk
n−1
X
k=0
Z 1 0
skds
= −
∞
X
n=1
tnHn Qn
k=1Hk
= −t 1 +
∞
X
n=2
tn−1 Qn−1
k=1Hk
!
= −tf (t) = 0
which contradicts the fact that I > 0. So we may conclude that f has no real zeros.
