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

n^{2} + · · · +Cm

n^{m} + o

1
n^{m}

,

where

C_{k} = (−1)^{k}

k

X

i=1

1
2^{i}

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 C_{1}= 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
2^{i−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) = 2x^{n−1} and Ek−1(x) = 2

k Bk(x) − 2^{k}Bk(x/2)
where B_{k}(x) is the k-th Bernoulli polynomial. Finally

C_{k} =(−1)^{k−1}E_{k−1}(1)

2 = (−1)^{k}E_{k−1}(0)

2 =(−1)^{k} B_{k}(0) − 2^{k}B_{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.