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γ + 1 2n− m X k=2 Bk k · 1 nk + o  1 nm

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Problem 11725

(American Mathematical Monthly, Vol.120, August-September 2013]) Proposed by Mher Safaryan (Armenia).

Let m be a positive integer. Show that, as n → ∞,

ln(2) −

n

X

k=1

(−1)k−1 k

= C1

n +C2

n2 + · · · +Cm

nm + o

 1 nm

 ,

where

Ck = (−1)k

k

X

i=1

1 2i

i

X

j=1

(−1)j i − 1 j − 1

 jk−1

for 1 ≤ k ≤ m.

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

By Euler-Maclaurin summation formula,

n

X

k=1

1

k = ln(n) + γ + 1 2n−

m

X

k=2

Bk k · 1

nk + o

 1 nm

 ,

and it follows that

ln(2) −

n

X

k=1

(−1)k−1 k

=

ln(2) −

n

X

k=1

1 k+

n/2

X

k=1

1 k

=

ln(2) − ln(n) − γ − 1 2n+

m

X

k=2

Bk

k · 1 nk

+ ln(n/2) + γ + 1 n−

m

X

k=2

Bk k · 2k

nk + o

 1 nm



= 1 2n−

m

X

k=2

(2k− 1)Bk

k · 1

nk + o

 1 nm

 .

Now C1= 1/2 and it suffices to show that Ck= −(2k− 1)Bk/k for k ≥ 2.

The (k − 1)-th Euler polynomial is given by

Ek−1(x) =

k

X

i=1

1 2i−1

i

X

j=1

(−1)j−1 i − 1 j − 1



(x + j − 1)k−1,

and it verifies the following known identities

Ek−1(x + 1) + Ek−1(x) = 2xn−1 and Ek−1(x) = 2

k Bk(x) − 2kBk(x/2) where Bk(x) is the k-th Bernoulli polynomial. Finally

Ck =(−1)k−1Ek−1(1)

2 = (−1)kEk−1(0)

2 =(−1)k Bk(0) − 2kBk(0)

k = −(2k− 1)Bk

k ,

where the last equality holds because Bk(0) = Bk and Bk = 0 if k ≥ 2 and k is odd. 

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