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# 1 2n · n−1 X k=0 2k+1 k+ 1

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Problem 11103

(American Mathematical Monthly, Vol.111, October 2004) Proposed by G. Galperin and H. Gauchman (USA).

Prove that for every positive integer n,

n

X

k=1

1 k nk =

1 2n−1

n

X

k = 1 k odd

n k

 k .

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We first note that

n

X

k=1

1 k nk =

n

X

k=1

1 n nk−1

−1

 = 1 n

n−1

X

k=0

1

n−1 k

 = 1 2n ·

n−1

X

k=0

2k+1 k+ 1.

We prove the last equality by induction (see also Problem 1682 of the Magazine). It holds for n= 1. Now take n > 1 and note that for k = 0, . . . , n − 2

1

n−1 k+1

 + 1

n−1 k

 =

n k+1



n−1 k+1



·

n−1 k

 =

n k+1·

n−1 k



n−1 k+1·

n−2 k



·

n−1 k

 = n n − 1·

1

n−2 k

 . Then summing over k and using the induction hypothesis we get

n−2

X

k=0

1

n−1 k+1

 +

n−2

X

k=0

1

n−1 k

 = n n − 1·

n−2

X

k=0

1

n−2 k

 = n 2n−1 ·

n−2

X

k=0

2k+1 k+ 1 that is

n−1

X

k=1

1

n−1 k

 +

n−2

X

k=0

1

n−1 k

 = 2

n−1

X

k=0

1

n−1 k

 − 1 − 1 = n 2n−1 ·

n−2

X

k=0

2k+1 k+ 1 and finally

1 n

n−1

X

k=0

1

n−1 k

 = 1 2n ·

n−2

X

k=0

2k+1 k+ 1 +1

n = 1 2n ·

n−1

X

k=0

2k+1 k+ 1. On the other hand

1 2n−1 ·

n

X

k = 1 k odd

n k

 k = 1

2n

n

X

k=1

n k

 Z 1

−1

xk−1dx= 1 2n

Z 1

−1

(1 + x)n− 1

x dx

= 1 2n

Z 2

0

yn− 1

y − 1 dy= 1 2n

n−1

X

k=0

Z 2

0

ykdy= 1 2n ·

n−1

X

k=0

2k+1 k+ 1. Hence

n

X

k=1

1 k nk =

1 2n ·

n−1

X

k=0

2k+1 k+ 1 = 1

2n−1

n

X

k = 1 k odd

n k

 k .



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