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1 + N X m=1 1 m! m(N −m+1) X r=m xr X 1≤k1,...,km k1+...+km=r 1 k1·k2

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Problem 11512

(American Mathematical Monthly, Vol.117, June–July 2010) Proposed by F. Holland (Ireland).

LetN be a nonnegative integer. For x ≥ 0 prove that

N

X

m=0

1 m!

N −m+1

X

k=1

xk k

!m

≥1 + x + . . . + xN.

Solution proposed by Paolo Perfetti and Roberto Tauraso,

Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let P (x) denote the polynomial on the left-hand side of the inequality

P (x) = 1 +

N

X

m=1

1 m!

N −m+1

X

k1=1

. . .

N −m+1

X

km=1

xk1+...+km k1·k2· · ·km =

= 1 +

N

X

m=1

1 m!

m(N −m+1)

X

r=m

xr X

1≤k1,...,km k1+...+km=r

1 k1·k2· · ·km.

It is easy to verify that the degree of P (x) is N (N + 2)/4 ≥ N if N is even and (N + 1)2/4 ≥ N if N is odd. Hence, since the coefficients of P (x) are nonnegative, it suffices to show that the coefficients of the monomials of degree 0, 1, . . . , N of P (x), are equal to 1, that is

1 n!

dn dxnP (x)

x=0= 1 n = 0, 1, 2, . . . , N.

For n = 0 and for any value of N it is trivial, so we assume that 1 ≤ n ≤ N . Therefore

1 n!

dn

dxnP (x) = 1 n!

N

X

m=1

1 m!

m(N −m+1)

X

r=m

xr−n X

1≤k1,...,km k1+...+km=r

r(r − 1)(r − 2) · · · (r − n + 1) k1·k2· · ·km

and finally we have that 1

n!

dn dxnP (x)

x=0=

N

X

m=1

1 m!

X

1≤k1,...,km k1+...+km=n

1 k1·k2· · ·km

= [xn]

N

X

m=1

1 m!logm

 1 1 − x



= [xn]

X

m=1

1 m!logm

 1 1 − x



= [xn]

 exp

 log

 1

1 − x



−1



= [xn] x 1 − x = 1 .



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