0 let An(x) :=Pn k=1 1 k Pk−1 j=0 n j
xj then n→∞lim An(x) (x + 1)n = ln  1 + 1 x  and lim n→∞n  (x + 1)nln  1 + 1 x  −An(x

Problem 12120

(American Mathematical Monthly, Vol.126, June-July 2019) Proposed by M. Bataille (France).

For positive integers n and k with n ≥ k, let a(n, k) =Pk−1 j=0

n j
3^j. (a) Evaluate

n→∞lim 1 4ⁿ

n

X

k=1

a(n, k) k . (b) Evaluate

n→∞lim n 4ⁿL −

n

X

k=1

a(n, k) k

!

where L is the limit in part (a).

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

Solution. We will prove a more general result: for x > 0 let An(x) :=Pn k=1

1 k

Pk−1 j=0

n

j
x^j then

n→∞lim

An(x) (x + 1)ⁿ = ln

 1 + 1

x



and lim

n→∞n



(x + 1)ⁿln

 1 + 1

x



−An(x)



= 1 x. Therefore, for x = 3, the limits are: (a) ln(4/3) and (b) 1/3.

We first show that the following recurrence holds

A0(x) = 0 , An(x) = (x + 1)An−1(x) + 1

n for n ≥ 1.

We have that

An(x) − (x + 1)An−1(x) =

n

X

k=1

1 k

k−1

X

j=0

n j



x^j−(x + 1)

n−1

X

k=1

1 k

k−1

X

j=0

n − 1 j

 x^j

= (x + 1)ⁿ−xⁿ

n +

n−1

X

k=1

1 k

k−1

X

j=1

n − 1 j − 1

 x^j−

n−1

X

k=1

1 k

k−1

X

j=0

n − 1 j

 x^j+1

= (x + 1)ⁿ−xⁿ

n +

n−1

X

k=1

1 k

k−1

X

j=1

n − 1 j − 1

 x^j−

n−1

X

k=1

1 k

k

X

j=1

n − 1 j − 1

 x^j

= (x + 1)ⁿ−xⁿ

n −

n−1

X

k=1

1 k

n − 1 k − 1



x^k= (x + 1)ⁿ−xⁿ

n − 1

n

n−1

X

k=1

n k

 x^k

= (x + 1)ⁿ−xⁿ

n −(1 + x)ⁿ−xⁿ−1

n = 1

n. By the above recurrence, it follows that An(x) = (x + 1)ⁿPn

k=1 1

k(x+1)^k and the first limit is

n→∞lim

An(x) (x + 1)ⁿ =

∞

X

k=1

1

k(x + 1)^k = ln 1 1 − _x+1¹

!

= ln

 1 + 1

x

 . Finally, for the last limit, we use the Stolz-Cesaro Theorem,

n→∞lim n



(x + 1)ⁿln

 1 + 1

x



−An(x)



= lim

n→∞

ln 1 +_x¹

−Pn k=1 1

k(x+1)^k 1

n(x+1)ⁿ SC= lim

n→∞

−_(n+1)(x+1)¹ n+1

1

(n+1)(x+1)ⁿ⁺¹ −_n(x+1)¹ n

= lim

n→∞

1

−1 +ⁿ⁺¹_n (x + 1) = 1 x.