Problem 12210
(American Mathematical Monthly, Vol.127, November 2020) Proposed by P. Bracken (USA).
Let x1= 1, and let
xn+1 =
√
xn+ 1
√xn
2
when n ≥ 1. For n ≥ 1, let
an= 2n + log(n) 2 − xn. Show that the sequence(an)n≥1converges.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. For n ≥ 1,
xn+1=
√
xn+ 1
√xn
2
= xn+ 1 xn
+ 2.
Let yn= xn− 2n then y1= −1, y2= 0, and yn+1− yn =yn+2n1 . Therefore, for n ≥ 3,
yn=
n−1X
k=2
1
yk+ 2k =⇒ 0 < yn≤
n−1X
k=2
1
2k =Hn−1− 1
2 .
Hence, for n ≥ 3,
an= log(n)
2 − yn= log(n)
2 −
n−1X
k=2
1 yk+ 2k
= 1
2(log(n) − Hn−1+ 1) +1 2
n−1X
k=2
1 k
1 − 1 1 +y2kk
n→∞−→ 1 − γ 2 +1
2 X∞ k=2
1 k
1 − 1 1 +y2kk
because
Hn = Xn k=1
1
k = log(n) + γ + o(1), and, for t ≥ 0, 0 ≤ 1 −1+t1 ≤ t implies
X∞ k=2
1 k
1 − 1 1 +y2kk
| {z }
≥0
≤ X∞ k=2
1 k· yk
2k ≤1 4
X∞ k=2
Hk−1− 1 k2
which is convergent since Hk−1k2−1 ∼log(k)k2 .
