The sequence (a n ) n ≥1 is defined by a 1 =1 and a n +1 = n/a n , for all inte- gers n ≥1 . Evaluate

1.3.10. The sequence (a n ) n ≥1 is defined by a 1 =1 and a n +1 = n/a n , for all inte- gers n ≥1 . Evaluate

W. W. Chao, Amer. Math. Monthly, Problem E 3356

Solution. We show that the limit is 2/ π + π /2. We first observe that

a n a n +1 = n and a n −1 a n = n −1, for all n ≥ 2. Subtracting these yields a n +1 −a n −1 =

a −1

n . Hence a −1 2 + ··· + a n =a n +a n +1 −a 1 −a 2 . It follows that

a −1

a n √ +1 = lim √ + √ . (1.9)

lim 1 +a −1 2 + ··· + a n

n →∞

n →∞

26 1 Sequences Since a n +1 = n/a n =a n −1 n /(n − 1), we can iterate to obtain

2 2n (n!) 2 (2n + 1)!

2 2n (n!) 2

√ By Stirling’s formula, we have a 2n +1 ∼ π /2 · 2n and a 2n +2 ∼ 2/ π · 2n.

Hence the two contributions of the right-hand side of (1.9) approach 2/ π and π /2. ⊓ ⊔

Elementary inequalities imply that the following recurrent sequence diverges like 2 √ n . As a consequence, a quite precise estimate for a high-order term is obtained.

1.3.11. Let (a n ) n ≥0

be the sequence defined by

a n =a n −1 + , for all n ≥ 1, a 0 = 5.

a n −1

Prove that 45 <a 1000 < 45.1 . Solution. We first observe that for all n ≥ 1,

It follows that a 2 >a 2 + 2n and, in particular, a 2 > 2025 = 45 n 2 0 1000 . For the reverse inequality, we write

2 1 1 2 1 1 =a n −1 +2+a n −1

a n − −1 a n By addition we obtain, for all n ≥ 1,

a 2 a 1000 1000 < 25 + 2000 +

and a straightforward computation shows that a 1000 < 45.1. ⊓ ⊔ Bounded sequences are not necessarily convergent. We give below a sufficient

condition that this happens. Monotonicity arguments again play a central role.

1.3.12. Let (a n ) n ≥0

be a bounded sequence of real numbers satisfying

a n +2 ≤ (a n +a n +1 ), for all n ≥ 0.

1.3 Recurrent Sequences 27

(a) Prove that the sequence (A n ) n ≥0 defined by A n = max{a n ,a n +1 } is convergent. (b) Deduce that the sequence (a n ) n ≥0 is convergent.

Solution. (a) We first observe that

a {a n ,a n +1 n } +2 ≤ =A n .

2 max

Since a n +1 ≤ max{a n ,a n +1 }=A n , it follows that max {a n +1 ,a n +2 }=A n +1 ≤A n , which shows that (A n ) n ≥0 is nonincreasing. Since (a n ) n ≥0 is bounded, it follows that the sequence (A n ) n ≥0 is also bounded, so it converges.

(b) By the boundedness of (a n ) n ≥0 , it follows that there exists a convergent sub- sequence (a n p ). Moreover, since (A n ) n ≥0 is nonincreasing, we can assume that

a n 1 ≥a n 2 ≥ ··· ≥ a n p ≥ ···. Set ℓ = lim p → ∞ a n p . Obviously, a n p ≥ ℓ. Set ℓ 1 = lim n → ∞ A n . We prove that ℓ≥ℓ 1 . Indeed, arguing by contradiction, it

follows that for any ε > 0, there exists a positive integer N ′ ( ε ) such that

a n p −1 <ℓ 1 + ε ,a n p <ℓ 1 − ε for all n p ≥N ′ ( ε ). On the other hand, a n p +1 ≤ 1/2(a n p +a n p −1 )<ℓ 1 and A n p = max{a n p ,a n p +1 }<ℓ 1 ,

which is impossible because the sequence (A n ) n ≥0 is nonincreasing and converges to ℓ 1 . By ℓ = lim p → ∞ a n p we obtain that for all ε > 0, there exists a positive integer N 1 =N 1 ( ε ) ∈ N such that a n p −ℓ< ε , for all n p ≥N 1 ( ε ). From ℓ 1 ≤ ℓ we obtain N 2 such that for all m ≥N 2 there exist n p and n p +1 satisfying a n p +1 ≤a m <a n p . Set N = max{N 1 ,N 2 }. Hence

a n −ℓ≤a n p −ℓ< ε , for all n ≥ N,

that is, ℓ = lim n → ∞ a n . An easier alternative argument is the following. Fix ε > 0. There is a positive integer N 1 such that for n ≥N 1 we have ℓ≤A n <ℓ+ ε . Hence a n ≤A n <ℓ+ ε . Suppose a n <ℓ−3 ε . Then a n +1 =A n <ℓ+ ε . Thus the recurrence gives

<ℓ− ε and a n +3 ≤

2 2 Hence A n +2 < ℓ, a contradiction. Thus for n ≥ N 1 we have ℓ −3 ε ≤a n < ℓ+ ε . Since