Problem 12012
(American Mathematical Monthly, Vol.124, December 2017) Proposed by O. Furdui and Alina Sintamarian (Romania).
Let k be a nonnegative integer. Find the set of real numbers x for which the power series
∞
X
n=k
n k
e− 1 − 1 1!− 1
2!− · · · − 1 n!
xn
converges, and determine the sum.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let
Fk(x) :=
∞
X
n=k
n k
e−
n
X
j=0
1 j!
xn. Note that by Taylor’s theorem, for n ≥ k, there is t ∈ (0, 1) such that
n k
e−
n
X
j=0
1 j!
xn
≤n k
et
(n + 1)!|x|n≤ e|x|k
k!(n + 1)· |x|n−k (n − k)!
and, by the Weierstrass M-Test, the power series Fk(x) converges for any real number x and it con- verges uniformly on any bounded subset in R (so we can interchange summation with differentiation).
Now, for k = 0,
F0(x) =
∞
X
n=0
e−
n
X
j=0
1 j!
xn=
∞
X
n=0
∞
X
j=n+1
1 j!
xn
=
∞
X
j=1
1 j!
j−1
X
n=0
xn
!
=
∞
X
j=1
1 j!
1 − xj 1 − x =
e− ex
1 − x if x 6= 1, e if x = 1.
By using the general Leibniz rule for the k-th derivative of a product, we get
D(k) e − ex 1 − x
= eD(k)((1 − x)−1) −
k
X
j=0
k j
D(k−j)(ex) · D(j)((1 − x)−1)
= k!e
(1 − x)k+1 − ex
k
X
j=0
k j
j!
(1 − x)j+1. Hence
Fk(x) = xk k!
∞
X
n=k
e−
n
X
j=0
1 j!
D(k)(xn) = xk
k! D(k)(F0(x))
=
xk (1 − x)k+1
e− ex
k
X
j=0
(1 − x)j j!
if x 6= 1, e
(k + 1)! if x = 1.
This is a more compact formula
Fk(x) = xkex
∞
X
j=0
(1 − x)j (j + k + 1)!.
