J. Numer. Anal. Approx. Theory, vol. 49 (2020) no. 1, pp. 45–53 ictp.acad.ro/jnaat
ON THE CONVERGENCE RATES OF THE PAIRS OF ADJACENT SEQUENCES
DOREL I. DUCA∗and ANDREI VERNESCU†
Abstract. In this paper we give a suitable definition for the pairs of adjacent (convergent) sequences of real numbers, we present some two-sided estimations which characterize the order of convergence to its limits of some of these se- quences and we give certain general explanations for its similar orders of conver- gence.
MSC 2010. 26D15, 30B10, 33F05, 40A05.
Keywords. sequence, limit, order of convergence, asymptotic scale, iterated limits.
1. INTRODUCTION
In the mathematical literature (e.g., [12, p. 112]) some of the most usual pairs of adjacent sequences are the following:
(a)an=endef==1 +n1n%e;1) bn=fndef==1 +n1n+1 &e;
(b)an=En
def== 1 + 1!1 +2!1 +. . .+n!1 %e; bn=En+n!n1 &e;
(c)an= 1 +√1
2 +√1
3 +. . .+ √1
n−2√
n+ 1%l;
bn= 1 +√1
2+ √1
3 +. . .+√1
n−2√ n&l
(where we have denoted byl the common limit of these last two sequences);
(d) an = 1 +n1n 1 +xn % e; bn = 1 +n1n 1 +yn & e (with x < 12 ≤y) (also see [16, p. 38]).
To these we can also add some other pairs deduced from [16, pp. 181–185]:
(e)an=1 +n1n+α %e; bn=1 +n1n+β &e (with α < 12 ≤β), (f) an=n!n−n−1/2en/exp12n1 %√
2π;
bn=n!n−n−1/2en/exp12n+1/41 &√
2π (see [11]),
∗ Faculty of Mathematics and Computer Science, Babe¸s-Bolyai University, Str.
Kog˘alniceanu, no. 1, Cluj Napoca, Romania, email: [email protected].
†Faculty of Sciences and Arts, Valahia University, Boulevard Unirii, no. 118, Tˆargovi¸ste, Romania, email: [email protected].
1) The oblique arrows show that the sequences tend increasing respectively decreasing to its limits. (N.A.)
(g)an= n+11 + n+21 +. . .+2n1 %ln 2;
bn= n+11 +n+21 +. . .+2n1 + 2n+11 &ln 2,
(h)an=1−n1n+1 %1/e; bn=1−n1n&1/e;
(i) an=Hn−ln(n+ 1)%γ; bn=γndef== Hn−lnn&γ
(where Hn= 1 +11+12 +13+. . .+1n is the harmonic sum of ordernand γ is the constant of Euler), see also [18, pp. 31–32, item 2.10];
(j)an(s) =ζn(s)−(n+1)1−s1−s %a(s);
bn(s) =ζn(s)−n1−s1−s &a(s) (with 0< s <1),
(where fors >0 we noteζn(s) = 1+21s+31s+. . .+n1s, fors∈(0,1),ζn(s)→ ∞ and a(s) = lim
n→∞an(s) = lim
n→∞bn(s)).
(k) an(s) =ζn(s) +(s−1)(n+1)1 s−1 %ζ(s);
bn(s) =ζn(s) +(s−1)n1 s−1 &ζ(s) (with s >1).
Fors >1,ζ(s) = lim
n→∞ζn(s) is the zeta-function of Riemann. For 0< s <1 the sequence (bn(s))n of (j) was considered by L. Euler (see [6], [7, pp. 112–
113]). The sequences of (k) appear in the proof of [29] of the known inequality:
1
(s−1)(n+ 1)s−1 < ζn−ζn(s)< 1 (s−1)ns−1 (see [8, vol. II, pp. 262–263] and [29]).
We must add now a basic standard pair of adjacent sequences namely (l) an=l−εn%l; bn=l+εn&l,
(where lis a fixed real number and (εn)n is a given sequence of positive num- bers which tends strictly decreasing to 0).
2. THE DEFINITION OF PAIRS OF ADJACENT SEQUENCES AND SOME EXPLANATIONS
All the pairs of convergent sequences previously mentioned satisfy, related to a given limit l, the conditions of Cantor Dedekind type in a strict form:
(1) a1< a2 < a3< . . . < an< . . . < bn< . . . < b3< b2< b1 (C-D1) (which express simultaneously two monotonicities and two boundednesses) and also the condition:
(2) lim
n→∞an= lim
n→∞bn=l.
[Of course, if we have the hypothesis (1) satisfied, the second condition of Cantor-Dedekind:
(3) lim
n→∞(bn−an) = 0 (C-D2)
implies that it exists an unique real numberl, such that we have the equality (2).]
But these conditions are not sufficient to assure that is suitable to call that the sequences (an)n and (bn)nconstitute a pair of adjacent sequences. We see a necessity to impose a certain condition of analytic relationship between the two sequences. Moreover, we also must put a condition of equal ,,velocity“ of tending to its common limit of the two sequences.
So we formulate the following
Definition 1. Two sequences (an)n and (bn)n are called to be adjacent related to a given limitlif its satisfy the conditions (1)and (2)and, moreover, it exists a nondegenerate interval I ⊂R and a function f :I×N→R, such that:
(i) For any t∈I, we have lim
n→∞f(t, n) =l;
(ii) It exists α, β ∈I, α < β, such that f(α, n) =an and f(β, n) =bn, for anyn∈N;
(iii) We have lim
n→∞
bn−l l−an = 1.
In the case of the pairs of sequences which were previous mentioned, we can takeI =Rand the functionsf :I×N→Rtogether with the valuesα and β can be obtained without difficulties. So, as example:
for (a): f(t, n) = (1 +n1)n+t;α= 0, β= 1;
for (b): f(t, n) =En+n!nt ; α= 0,β = 1;
for (c);f(t, n) = 1 +√12 +√1
3 +. . .+√1n−2pn+ (1−t);α= 0, β = 1;
for (d): f(t, n) =en 1 +nt; α =x < 12 ≤y=β;
for (e): f(t, n) =1 +n1n+t; α < 12 ≤β;
for (f): f(t, n) =n!n−n−1/2exp n−12n+t1 ; α= 0,β = 14; for (g): f(t, n) = n+11 +n+21 +. . .+2n1 +2n+1t ; α= 0, β= 1;
for (h): f(t, n) =1−n1n+(1−t); α= 0,β = 1;
for (i): f(t, n) =Hn−ln(n+ (1−t)); α= 0, β= 1;
for (j): f(t, n) = 1 +21s +31s +. . .+n1s −(n+(1−t))1−s 1−s; α= 0,β = 1;
for (k): f(t, n) = 1 + 21s +31s +. . .+n1s −(s−1)(n+(1−t))1 s−1; α= 0, β= 1;
for (l): f(t, n) =l+εn·t; α=−1,β = 1.
For x = 0 and y = 1 in (d) we obtain (a); For α = 0 andβ = 1 in (e) we obtain again (a). (So we see that, for a given pair of adjacent sequences, the functionf can be not unique.) For s= 12 in (j) we obtain (c).
Also a certain convergent and monotonic sequence admits no an unique adjacent pair; if (an)n and (bn)n are as in the definition, then, for any β1> β (β1 ∈I), the sequence n 7→ f(β1, n) is a pair of (an)n; also, for any α1 < α (α1 ∈ I), the sequence n 7→ f(α1, n) is a pair of (bn)n. To illustrate the necessity of the condition (iii) of our definition, consider the sequences of general term:
xn= 1 +1 2+ 1
3+. . .+ 1
n−1 + 1
3n−lnn;
yn= 1 +1 2 +1
3 +. . .+ 1
n−1 + 1
2n−lnn (see [30]).
Both these sequences converge to the Euler’s constantγ; they are obtained modifying not the logarithm (as in [4], [13]), but the last term of the harmonic sum Hn. The sequence (xn)n is strictly increasing and the sequence (yn)n is strictly decreasing. Also, we can choose the function f : R×N → R, f(t, n) =Hn−1+nt −lnnand the valuesα= 13 and β= 12. But the sequences (xn)n and (yn)n are not adjacent because the condition (iii) from definition is not satisfyied; we have: lim
n→∞
yn−γ
γ−xn = 0. [More precisely, we observe that
n→∞lim n2(γ −yn) = 121 and lim
n→∞n(γ −xn) = 16. All the three last results can be obtained using the so called lemma of Stolz-Ces`aro for the case 00 (see [9, p. 54] and [17]). In [30] the two-estimate 12(n+1)1 2 < γ−yn< 12n12 is proved.
The explanation consists in the asymptotic expansion of Hn, namely:
Hn= lnn+γ+ 1 2n+ 1
12n2 +. . .]
3. A SPECIAL CASE
For some strictly monotonic, convergent sequences the finding of an adjacent pair may appear sometime unexpected. So, the sequence (bn)nof general term bn= √n
n, which tends to 1 and is strictly decreasing (forn≥3) can not have as adjacent pair the sequence of general terman= n+1√
n, because this, although tends to 1, is also strictly decreasing (for n≥3). A possible adjacent pair of (bn)n is the sequence an = n√1
n, (n≥ 3) which tends strictly increasing to 1;
in this case I =R and f(t, n) = (√n
n)t with the values α = −1, β = 1; the conditiion (iii) is also satisfyied.
4. SOME TWO-SIDED ESTIMATIONS WHICH DESCRIBE THE FIRST ORDER OF CONVERGENCE
The order of convergence of the majority of the sequences of the Section 1 can be described by certain pairs of two-sided estimations. So, we have:
• for (a):
(4) e
2n+ 2 <e−
1 + 1 n
n
< e
2n+ 1; (see [16, p. 38], [27])
(40) e
2n+ 1 <
1 + 1
n n+1
−e< e
2n; (see [25])
• for (c):
(5) 1
2√
n+ 1 < a−an< 1 2√
n;
(50) 1 2√
n+ 1 < bn−a < 1 2√
n, (see [28])
• for (g):
(6) 1
4n+α <ln 2−an< 1
4n+ 1 (α >1); (see [31])
(60) 1
4n+β < bn−ln 2< 1
4n+ 3 (β >3);
• for (h):
(7) 1
2ne < 1 e −
1−1
n n+1
< 1
(2n−1)e; (see [14], [15])
(70) 1
(2n−1)e <
1− 1
n n
−1
e < 1
(2n−2)e (n≥2) (see [14], [15]),
• for (i):
(8) 1
2n+ 1 <(Hn−lnn)−γ < 1
2n; (see [26])
(80) 1
2n+ 2 < γ−(Hn−ln(n+ 1))< 1 2n+ 1. So, for the examples (5) and (50), we havea−an=O(√1
n),bn−a=O(√1
n), and for all the other the median term of the two sided estimations is O(n1) (recall that αn = O(βn) if it exists a constant C > 0 and n0 ∈ N such that
|αn| ≤C|βn|, for alln≥n0).
5. THE FIRST ITERATED LIMIT
We recall now some basic facts of the asymptotic analysis; we present its directly for the sequences, i.e. for the functions of natural variable, in a neighborhood of the unique accumulation point of the domain of definitionN, of the sequences, namely ∞.
If (xn)n and (yn)n are two sequences we call that these are asymptotic equivalent (and we write xn ∼ yn) if lim
n→∞xn/yn exists and is a finite and different of 0 number.
Consider now a convergent sequence (xn)n = (x(n))n with the limit l. A sequence (uk)k of functions of natural variablen,uk =uk(n), k= 0,1,2, . . ., with n ∈ N is called to be an asymptotic scale for the given sequence (xn)n if it exists a sequence of real numbers (lk)k, wherel0 =l, such that, for any k∈N, we have:
xn∼l0+l1u1(n) +l2u2(n) +. . .+lkuk(n),
where, for everyj = 0,1, . . . , k−1, we haveuj+1=o(uj),i.e. lim
n→∞
uj+1(n) uj(n) = 0.
The coefficients l1,l2, l3,. . . , are also called the iterated limits of sequence (xn)n(with respect to the asymptotic scale (uk)k). (See the expository books of Copson [2], De Bruijn [3], Erdely [5], van der Corput [19]–[24].)
All our inequalities of the section 3 implies that, for two adjacent sequences, the first iterated limit (related to the same function n 7→ u1(n) of a given asymptotic scale of functions of natural variable uk = uk(n), k= 0,1,2, . . .) is the same. So, we have:
• for (a):
(9) lim
n→∞ne− 1 +1nn= e2;
(90) lim
n→∞n 1 +1nn+1−e= e2;
• for (c):
(10) lim
n→∞
√n(l−an) = 12; (100) lim
n→∞
√n(bn−l) = 12;
• for (g):
(11) lim
n→∞n(ln 2−an) = 14; (110) lim
n→∞n(bn−ln 2) = 14;
• for (h):
(12) lim
n→∞n
1
e−1−n1n+1
= 2e1;
(120) lim
n→∞n1−n1n−1e= 2e1;
•for (i):
(13) lim
n→∞n(γ−(Hn−ln(n+ 1))) = 12;
(130) lim
n→∞n((Hn−lnn)−γ) = 12. We can present an explanation of this fact.
Proposition 2. If (an)n and (bn)n are two adjacent sequences, then, for any asymptotic scale (uk)k (the same for both the sequences), its first iterated limits are equal.
Proof. Letl1andλ1be the first iterated limits of (an)nand (bn)n. Suppose, ad absurdum, thatl1 6=λ1. Then we have:
n→∞lim bn−l l−an
=
n→∞lim u1(n)(bn−l)
n→∞lim u1(n)(l−an) = λ1 l1
6= 1, a contradiction! Therefore l1=λ1.
This shows for what, in all the previous two-sided estimations, the principal
coefficient must be the same.
6. SOME RESULTS
We can give now some general results concerning the order of convergence of the sequences of adjacent pairs.
Proposition3. Let(an)nand(bn)nbe two adjacent sequences. Iff admits a partial derivative respecting the variable t and, for any n ∈ N, it exists mn, Mn∈R, mn< Mn such that:
mn≤ ∂f
∂t(t, n)≤Mn (for anyt∈I).
Then:
(β−α)mn≤bn−an≤(β−α)Mn (for anyn∈N).
Proof. Letn ∈N be fixed. So we can consider the function t7→ f(t, n) as a function of one variablet andf0(t, n) = ∂f∂t(t, n). In view of the mean value theorem of Lagrange, there existsτ ∈(α, β) such that:
bn−an=f(β, n)−f(α, n) = ∂f
∂t(τ, n)(β−α),
which gives our conclusion.
Corollary 4. In the same hypotesis we have:
l−an<(β−α)Mn, bn−l <(β−α)Mn
(because l−an< bn−an and bn−l < bn−an).
7. THE CASE OF CONTINUATION
For many sequences of real numbers (an)n≥1 it is possible to find a function g : [1,∞) → R such that g(n) =an, for any n∈ N, called a continuation of the sequence on the positive real axis.
Two of nontrivial examples are the factorial and the harmonic sum.
So, for an = n!, the continuation to [0,∞) is made by the Γ function of Euler,
Γ(x) = Z ∞
0
tx−1e−tdt, x≥1.
As Γ(x+ 1) = xΓ(x) and Γ(1) = 1, we have Γ(n+ 1) = n!. Therefore the functiong: [0,∞)→R,g(x) =R0∞txe−tdt is the continuation of the factorial on the positive axis [0,∞).
The functiom Γ is also logarithmically-convex. A beautiful theorem of Bohr and Mollerup characterizes completely the function: if f : [0,∞) → [0,∞) satisfies the functional equation f(x+ 1) = xf(x), is logarithmically-convex and f(1) = 1, thenf = Γ.
For the sequence (Hn)n≥1 of the harmonic sums, the continuation function is H : (0,∞)→ R, defined by the equality H(x) =ψ(x+ 1)−γ, where ψ is the logarithmic derivative of Γ, ψ(x) = Γ0(x)/Γ(x), and γ is the constant of Euler.
Suppose now that the adjacent sequences (an)n and (bn)n admits continu- ations on [1,∞). In this case the function f :I ×N → R admits a natural continuation onI×[1,∞).
Proposition 5. Let (an)n≥1 and (bn)n≥1 be two sequences which admit continuations to [1,∞) and for which it exists a function f which satisfies the conditions (i) and (ii) of the definition. Iff admits a partial derivative and it exists in R lim
x→x0
∂f
∂x(α, x) = lim
x→x0
∂f
∂x(β, x) 6= 0, then the sequences (an)n and (bn)n are adjacent.
Proof. The sequences of general termbn−l, respectivelyl−anhas strictly positive terms and tend decreasing to zero. We obtain:
x→xlim0
(bn+1−l)−(bn−l)
(an+1−l)−(an−l) = lim
x→x0
f(β,n+1)−f(β,n)
f(α,n+1)−f(α,n) = lim
x→x0
∂f
∂x(β, ν1)
∂f
∂x(α, ν2)
where ν1 and ν2 are contained in the interval (n, n+ 1) and tend to ∞ when n → ∞. From the hypotesis, the last limit is equal to 1. According to the lemma of Stolz-Ces`aro for the case 00 (see [9, p. 56], [17]) we have the
conclusion.
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Received by the editors: January 9, 2020; accepted: May 19, 2020; published online:
August 11, 2020.
