(b) Find lim n→∞ ln  e n + a

Problem 11851

(American Mathematical Monthly, Vol.122, June-July 2015) Proposed by D. M. B˘atinet,u-Giurgiu, and N. Stanciu (Romania).

For real a and b and integer n ≥ 1, let γn(a, b) = − ln(n + a) +

n

X

k=1

1 k + b. (a) Prove that γ(a, b) = lim

n→∞γn(a, b) exists and is finite.

(b) Find

lim

n→∞ ln

 e n + a

 +

n

X

k=1

1

k + b− γ(a, b)

!ⁿ .

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

We have that if x > 0 then

Ψ(x) = −γ +

∞

X

k=1

 1

k − 1

x − 1 + k



= ln(x) − 1

2x+ O(1/x²), which implies that (b should be not a negative integer)

γ_n(a, b) + Ψ(b + 1) = − ln(n + a) +

n

X

k=1

1

k + b+ Ψ(b + 1)

= − ln(n + a) + Ψ(b + n + 1)

= − ln(n + a) + ln(b + n + 1) − 1

2(b + n + 1) + O(1/n²)

= ln b + n + 1 n + a



− 1

2(b + n + 1) + O(1/n²)

= b − a +¹₂

n + O(1/n²) Therefore γ(a, b) = −Ψ(b + 1). Moreover,

ln

 e n + a

 +

n

X

k=1

1

k + b− γ(a, b) = 1 − ln(n + a) +

n

X

k=1

1

k + b+ Ψ(b + 1)

= 1 + b − a +¹₂

n + O(1/n²).

and finally we get

n→∞lim ln

 e n + a

 +

n

X

k=1

1

k + b− γ(a, b)

!n

= exp



b − a +1 2

 .