Problem 11597
(American Mathematical Monthly, Vol.118, October 2011)
Proposed by Michel Bataille (France).
Let f (x) = x/ log(1 − x). Prove that for 0 < x < 1,
∞
X
n=1
xn(1 − x)n
n! f(n)(x) = −1 2xf (x).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
The function f has a power expansion at 0 given byP∞
k=1akxk for x ∈ (0, 1). Moreover, for m ≥ 1,
[xm]
∞
X
n=1
xn(1 − x)n
n! f(n)(x) = [xm]
∞
X
n=1
xn
n
X
j=0
n j
∞ X
k=n
k n
akxk−n
=
m
X
n=1 m
X
k=n
n
m − k
(−1)m−kk n
ak
=
m
X
k=1
(−1)m−kak k
X
n=1
n
m − k
k n
=
m
X
k=1
(−1)m−kak
22km
k m − k
−
0 m − k
.
Hence, the identity holds as soon as
m
X
k=1
(−1)m−kak
22km
k m − k
−
0 m − k
= [xm] − 1
2xf (x) = −1 2am−1,
that is if and only if
m
X
k=1
k m − k
(−4)kak = (−2)mam+ (−2)m−1am−1.
SincePm k=1
k
m−k[tk]g(t) = [tm]g(t(1 + t)), it follows that
m
X
k=1
k m − k
(−4)kak = [tm]f (−4t) = [tm] −4t(1 + t) log(1 + 4t + t2)
= [tm]f (−2t) + [tm−1]f (−2t) = (−2)mam+ (−2)m−1am−1
and the proof is complete.
