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

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

n^m + o

 1 n^m

 ,

where

C_k = (−1)^k

k

X

i=1

1 2ⁱ

i

X

j=1

(−1)^j i − 1 j − 1

 j^k−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

B_k k · 1

n^k + o

 1 n^m

 ,

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 n^k

+ ln(n/2) + γ + 1 n−

m

X

k=2

B_k k · 2^k

n^k + o

 1 n^m

    

= 1 2n−

m

X

k=2

(2^k− 1)Bk

k · 1

n^k + o

 1 n^m

 .

Now C₁= 1/2 and it suffices to show that C_k= −(2^k− 1)B_k/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) − 2^kBk(x/2)
where B_k(x) is the k-th Bernoulli polynomial. Finally

C_k =(−1)^k−1E_k−1(1)

2 = (−1)^kE_k−1(0)

2 =(−1)^k B_k(0) − 2^kB_k(0)

k = −(2^k− 1)B_k

k ,

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