extending tests for convergence of number...

Extending tests for convergence of number series

Elijah Liflyand, Sergey Tikhonov, and Maria Zeltser

September, 2012

Liflyand et al Extending monotonicity September, 2012 1 / 28

Goals and History

Main Goal:

Relax the monotonicity assumption for the sequence of terms of the series.

In various well-known tests for convergence/divergence of numberseries

∞∑k=1

ak, (1)

with positive ak, monotonicity of the sequence of these ak is the basicassumption.

Such series are frequently called monotone series.

Tests by Abel, Cauchy, de la Vallee Poussin, Dedekind, Dirichlet, duBois Reymond, Ermakov, Leibniz, Maclaurin, Olivier, Sapogov,Schlomilch are related to monotonicity.

Goals and History

Main Goal:

∞∑k=1

ak, (1)

Goals and History

Main Goal:

∞∑k=1

ak, (1)

Goals and History

Main Goal:

∞∑k=1

ak, (1)

References

Books:

T.J.I’a Bromwich, Introduction to the Theory of Infinite Series,MacMillan, New York, 1965.

K. Knopp, Theory and Application of Infinite Series, Blackie& SonLtd., London-Glasgow, 1928.

G.M. Fichtengolz, Infinite Series: Rudiments, Gordon and Breach,New York, 1970.

D.D. Bonar and M.J. Khoury, Real Infinite Series, MAA, Washington,DC, 2006.

References

Books:

References

Books:

References

Books:

References

Books:

In its initial form the Maclaurin-Cauchy integral test reads as follows:

Consider a non-negative monotone decreasing function f defined on[1,∞). Then the series

∞∑k=1

f(k) (2)

converges if and only if the integral

∫ ∞1

f(t) dt

is finite. In particular, if the integral diverges, then the series diverges.

Ermakov’s test.

In its simplest form, it is given as follows.

Let f be a continuous (this is not necessary) non-negative monotonedecreasing function for t > 1.

If for t large enough f(et)et/f(t) ≤ q < 1, then series (2) converges,

while if f(et)et/f(t) ≥ 1, then series (2) diverges.

In fact, it is known that one can get a family of tests by replacing et

with a positive increasing function ϕ(t) satisfying certain properties.

We will extend such a more general assertion.

Ermakov’s test.

The well known Cauchy condensation test states that

If {ak} is a positive monotone decreasing null sequence, then theseries (1) and

∞∑k=1

converge or diverge simultaneously.

This assertion is a partial case of the following classical result due toSchlomilch.

Let (1) be a series whose terms are positive and non-increasing,

∞∑k=1

and let

u0 < u1 < u2 < ... be a sequence of positive integers such that

∆uk∆uk−1

≤ C.

Then series (1) converges if and only if the series

∞∑k=1

∆ukauk =∞∑k=1

(uk+1 − uk)auk

converges.

In the theory of monotone series there are statements on the behaviorof its terms. Such is Abel–Olivier’s kth term test:

Let {ak} be a positive monotone null sequence. If series (1) isconvergent, then kak is a null sequence.

and let

∆uk∆uk−1

≤ C.

∞∑k=1

∆ukauk =

∞∑k=1

(uk+1 − uk)auk

converges.

and let

∆uk∆uk−1

≤ C.

∞∑k=1

∆ukauk =

∞∑k=1

(uk+1 − uk)auk

converges.

and let

∆uk∆uk−1

≤ C.

∞∑k=1

∆ukauk =

∞∑k=1

(uk+1 − uk)auk

converges.

Sapogov’s test reads as follows.

If {bk} is a positive monotone increasing sequence, then the series

∞∑k=1

(1− bkbk+1

as well as

∞∑k=1

bk− 1)

converges if the sequence {bk} is bounded and diverges otherwise.

The tests of Dedekind and of du Bois Reymond are united as the nextassertion.

Sequence of bounded variation – {ak} ∈ BV :∑|ak+1 − ak| <∞.

Let {ak} and {bk} be two sequences.

(i) If {ak} ∈ BV , {ak} is a null sequence, and the sequence of partial

sums of∑k

bk is bounded, then the series∞∑k=1

akbk is convergent.

(ii) If {ak} ∈ BV and∑k

bk is convergent, then the series∞∑k=1

akbk is

convergent.

Two well-known and widely used corollaries of this test -

Dirichlet’s and Abel’s tests - involve monotone sequences.

sums of∑k

akbk is convergent.

akbk is

convergent.

sums of∑k

akbk is convergent.

akbk is

convergent.

sums of∑k

akbk is convergent.

akbk is

convergent.

sums of∑k

akbk is convergent.

akbk is

convergent.

The first one is as follows.

Let {ak} be a monotone null sequence and {bk} be a sequence suchthat the sequence of its partial sums is bounded. Then the series∞∑k=1

akbk is convergent.

One of corollaries of this test is the celebrated Leibniz test:

Let {ak} be a monotone null sequence. Then the series∞∑k=1

(−1)kak

is convergent.

Further, Abel’s test reads as follows.Let {ak} be a bounded monotone sequence and

bk a convergent

series. Then the series∞∑k=1

akbk is convergent.

(−1)kak

is convergent.

bk a convergent

akbk is convergent.

(−1)kak

is convergent.

bk a convergent

akbk is convergent.

(−1)kak

is convergent.

bk a convergent

akbk is convergent.

Classes of sequences and functions

Let us introduce certain classes of functions and sequences more generalthan monotone.

Definition

We call a non-negative null (that is, tending to zero at infinity) sequence{ak} weak monotone, written WMS, if for some positive absolute constantC it satisfies

ak ≤ Can for any k ∈ [n, 2n].

To introduce a counterpart for functions, we will assume all functionsto be defined on (0,∞), locally of bounded variation, and vanishingat infinity.

Definition

We say that a non-negative function f defined on (0,∞), is weakmonotone, written WM, if

f(t) ≤ Cf(x) for any t ∈ [x, 2x].

Setting ak = f(k) in this case, we obtain {ak} ∈WMS.

Clearly, in these definitions 2n and 2x can be replaced by [cn] (where[a] denotes the integer part of a) and cx, respectively, with somec > 1 and another constant C.

In some problems one should consider a smaller class than WMS.

Definition

Denote ∆ak = ak+1 − ak.

Definition. A positive null sequence {ak} is called general monotoneif it satisfies

2n∑k=n

|∆ak| ≤ Can,

for any n and some absolute constant C.

The class of quasi-monotone sequences,

that is, {ak} such that there exists τ > 0 so that k−τak ↓,is a proper subclass of GMS.

One of the simple basic properties of GMS is GMS (WMS.

2n∑k=n

|∆ak| ≤ Can,

2n∑k=n

|∆ak| ≤ Can,

2n∑k=n

|∆ak| ≤ Can,

that is, {ak} such that there exists τ > 0 so that k−τak ↓,

is a proper subclass of GMS.

2n∑k=n

|∆ak| ≤ Can,

2n∑k=n

|∆ak| ≤ Can,

A similar function class:

Definition

We say that a non-negative function f is general monotone, GM, if for allx ∈ (0,∞) ∫ 2x

x|df(t)| ≤ Cf(x).

We just remark that if f(·) is a GM function,

then for ak = f(k) there holds {ak} ∈ GMS.

And of course GM (WM .

Definition

x|df(t)| ≤ Cf(x).

Definition

x|df(t)| ≤ Cf(x).

Definition

x|df(t)| ≤ Cf(x).

Extension of ”monotone” tests

An extension of the Maclaurin-Cauchy integral test reads as follows:

Theorem

Let f be a WM function. Then the series

∞∑k=1

and integral ∫ ∞1

f(t) dt

An extension of Ermakov’s test:

Theorem

Let f be a WM function and let ϕ(t) be a monotone increasing, positivefunction having a continuous derivative and satisfying ϕ(t) > t for all tlarge enough. If for t large enough

f(ϕ(t))ϕ′(t)

f(t)≤ q < 1,

then series (2) converges, while if

f(ϕ(t))ϕ′(t)

f(t)≥ 1,

then series (2) diverges.

Theorem

Let {uk} be an increasing sequence of positive numbers such thatuk+1 = O(uk) and uk →∞.

Let f be a WM function. Then both series

∞∑k=1

f(uk)∆uk and∞∑k=1

f(uk+1)∆uk

converge (or diverge) with∫∞1 f(t) dt.

Let {ak} be a WMS. Then (1) converges if and only if the series

∞∑k=1

∆ukauk =∞∑k=1

(uk+1 − uk)auk

converges.

Theorem

∞∑k=1

f(uk)∆uk and

∞∑k=1

f(uk+1)∆uk

∞∑k=1

∆ukauk =∞∑k=1

(uk+1 − uk)auk

converges.

Theorem

∞∑k=1

f(uk)∆uk and

∞∑k=1

f(uk+1)∆uk

∞∑k=1

∆ukauk =

∞∑k=1

(uk+1 − uk)auk

converges.

An extension of Abel–Olivier’s kth term test:

Theorem

Let {ak} be a WMS. If series (1) converges, then kak is a null sequence.

Analyzing one Dvoretzky’s result and its proof,

we obtain the following statement.

Theorem

If {ak}, {bk} ∈WMS, and the series are convergent and divergent,respectively, then for every M > 1 there exist infinitely many Rj ,Rj →∞, such that for all k with Rj ≤ k ≤MRj , we have ak < bk.

Theorem

Our next result is ”dual” to the above Schlomilch-type extensions.

We recall that the increasing sequence {uk} is called lacunary ifuk+1/uk ≥ q > 1.

A more general class of sequences is the one in which each sequencecan be split into finitely-many lacunary sequences. In the latter casewe will write {uk} ∈ Λ.

This is true if and only if

k∑j=1

uj ≤ Cuk.

In different terms, {uk} ∈ Λ is true if and only if there exists r ∈ Nsuch that

uk+ruk≥ q > 1, k ∈ N.

k∑j=1

uj ≤ Cuk.

uk+ruk≥ q > 1, k ∈ N.

k∑j=1

uj ≤ Cuk.

uk+ruk≥ q > 1, k ∈ N.

k∑j=1

uj ≤ Cuk.

uk+ruk≥ q > 1, k ∈ N.

k∑j=1

uj ≤ Cuk.

uk+ruk≥ q > 1, k ∈ N.

We denote ∆auk := auk − auk+1.

Proposition. Let {ak} be a non-negative WMS, and let a sequence {uk}be such that {uk} ∈ Λ and uk+1 = O(uk).Then the series (1),

∞∑k=1

uk|∆auk |,

∞∑k=1

General monotonicity

It is also possible to get equiconvergence results for an important caseun = n, where the lacunarity can no more help.

We proceed to a smaller class than WMS. Indeed, assuming generalmonotonicity of the sequences, we prove the following result.

Proposition. Let {ak} be a GMS. Then series (1) and

k|∆ak|

A similar result for functions:

Proposition. Let f be a GM function. Then the integrals∫∞1 f(t) dt and

∫∞1 t|df(t)| converge or diverge simultaneously.

k|∆ak|

Negative type results

We now proceed to a group of tests where

extending monotonicity to weak monotonicity fails in that or anothersense.

Sapogov type test cannot be true if {bk} (as well as {1/bk}) is WMS.

A counterexample can be constructed against generalization of Abel’stest.

As for extending the Leibniz test, it cannot hold without additionalassumption of the boundedness of variation:

just take ak = 1/ ln k everywhere except n = 2k where an = 2/ lnn.

Wider classes

We shall show that WMS is, in a sense, the widest class for which suchtests are still valid.

An immediate natural extension of WMS is the class defined by

ak ≤ Ck∑

or a bit more general

ak ≤ C[ck]∑

n=[k/c]

for some c > 1.

Wider classes

ak ≤ Ck∑

ak ≤ C[ck]∑

n=[k/c]

for some c > 1.

Wider classes

ak ≤ Ck∑

ak ≤ C[ck]∑

n=[k/c]

for some c > 1.

Wider classes

The principle difference between WMS and these classes is that the lattertwo allow certain amount of zero members, unlike WMS that forbid evena single zero, i.e., an0 = 0 implies an = 0 for n ≥ n0.

Putting zeros on certain positions, say k = 2n, we easily construct acounterexample to show that the Cauchy condensation test cannot bevalid for these classes nor its extensions.

In a similar way, just letting a function f to take non-zero values onlyclose to integer points, one sees that the Maclaurin-Cauchy integraltest may fail as well.

In conclusion, note that WMS is a subclass of the broadly used∆2-class,

Wider classes

that is, the one

for which the doubling condition a2k ≤ Cak holds for each k ∈ N.We observe that assuming the doubling condition by no means canguarantee the above tests to be extended.To illustrate this, the easiest way is to consider Cauchy’scondensation test.Take {ak} such that

{2−k, k 6= 2n;n−2, k = 2n.

This sequence is doubling but not WMS.Obviously,

∞∑n=1

2na2n =

∞∑n=1

diverges

Wider classes

that is, the one

for which the doubling condition a2k ≤ Cak holds for each k ∈ N.

We observe that assuming the doubling condition by no means canguarantee the above tests to be extended.To illustrate this, the easiest way is to consider Cauchy’scondensation test.Take {ak} such that

{2−k, k 6= 2n;n−2, k = 2n.

∞∑n=1

2na2n =

∞∑n=1

diverges

Wider classes

that is, the one

for which the doubling condition a2k ≤ Cak holds for each k ∈ N.We observe that assuming the doubling condition by no means canguarantee the above tests to be extended.

To illustrate this, the easiest way is to consider Cauchy’scondensation test.Take {ak} such that

{2−k, k 6= 2n;n−2, k = 2n.

∞∑n=1

2na2n =

∞∑n=1

diverges

Wider classes

that is, the one

for which the doubling condition a2k ≤ Cak holds for each k ∈ N.We observe that assuming the doubling condition by no means canguarantee the above tests to be extended.To illustrate this, the easiest way is to consider Cauchy’scondensation test.

Take {ak} such that

{2−k, k 6= 2n;n−2, k = 2n.

∞∑n=1

2na2n =

∞∑n=1

diverges

Wider classes

that is, the one

{2−k, k 6= 2n;n−2, k = 2n.

∞∑n=1

2na2n =

∞∑n=1

diverges

Wider classes

that is, the one

{2−k, k 6= 2n;n−2, k = 2n.

This sequence is doubling but not WMS.

Obviously,

∞∑n=1

2na2n =

∞∑n=1

diverges

Wider classes

that is, the one

{2−k, k 6= 2n;n−2, k = 2n.

∞∑n=1

2na2n =

∞∑n=1

divergesLiflyand et al Extending monotonicity September, 2012 25 / 28

Overview

The main results may be summarized as the following statement.

Theorem. Let f(·) ∈WM . Then the following series and integralsconverge or diverge simultaneously:∫ ∞

1f(t) dt;

(uk+1 − uk)f(uk), provided uk ↑, uk+1 = O(uk);

ukf(uk),∑k

uk|f(uk+1)− f(uk)|, withuk lacunary, uk+1 = O(uk);

k|f(k + 1)− f(k)|,∫ ∞1

t |df(t)|, provided f(·) is GM .

Overview

Theorem. Let f(·) ∈WM . Then the following series and integralsconverge or diverge simultaneously:

∫ ∞1

f(t) dt;∑k

ukf(uk),∑k

k|f(k + 1)− f(k)|,∫ ∞1

Overview

1f(t) dt;

ukf(uk),∑k

k|f(k + 1)− f(k)|,∫ ∞1

Overview

1f(t) dt;

ukf(uk),∑k

k|f(k + 1)− f(k)|,∫ ∞1

Overview

1f(t) dt;

ukf(uk),∑k

k|f(k + 1)− f(k)|,∫ ∞1

Overview

1f(t) dt;

ukf(uk),∑k

k|f(k + 1)− f(k)|,∫ ∞1

Example

Besov space:

For k > r, θ, r > 0 and p ≥ 1

‖f‖Brp,θ =

t−rθ−1ωθk(f ; t)pdt

Equivalent form:

‖f‖Brp,θ =

∞∫1

trθ+1ωθk(f ; 1/t)pdt

Example

Besov space:

For k > r, θ, r > 0 and p ≥ 1

‖f‖Brp,θ =

Equivalent form:

‖f‖Brp,θ =

∞∫1

Example

Besov space:

For k > r, θ, r > 0 and p ≥ 1

‖f‖Brp,θ =

Equivalent form:

‖f‖Brp,θ =

∞∫1

Example

Since ωk(f ; t)p ↑, the function under the integral sign is WM .

Applying generalized Maclaurin-Cauchy and Cauchy condensationtests, we obtain

convenient equivalent form:

( ∞∑ν=0

[2rνωk(f ; 2−ν)p

]θ)1/θ

Example

( ∞∑ν=0

]θ)1/θ

Example

( ∞∑ν=0

]θ)1/θ

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