Summability characterizations of positive sequences
aa r X i v : . [ m a t h . C A ] J a n Summability characterizations of positivesequences
Douglas Azevedo and Thiago P. de AndradeJanuary 12, 2021
Abstract
In this paper, we propose extensions for the classical Kummer test,which is a very far-reaching criterion that provides sufficient and neces-sary conditions for convergence and divergence of series of positive terms.Furthermore, we present and discuss some interesting consequences andexamples such as extensions of the Olivier’s theorem and Raabe, Bertrandand Gauss’s test.
The Kummer’s test is an advanced theoretical test which provides necessaryand sufficient conditions that ensures convergence and divergence of series ofpositive terms. Below we present the statement of this result. Its proof andsome additional historical background may be found in [1, 4, 9].
Theorem 1. (Kummer’s test) Consider the series P a n where { a n } is a se-quence of positive real numbers. ( i ) The series P a n converges if and only if that there exist a sequence { q n } ,a real number c > and an integer N ≥ for which q n a n a n +1 − q n +1 ≥ c, n ≥ N. ( ii ) The series P a n diverges if, and only if there exist a sequence { q n } andan integer N ≥ for which P q n is a divergent series and q n a n a n +1 − q n +1 ≤ , n ≥ N. Besides providing an extremely far-reaching characterization of convergenceand divergence of series with positive terms, the importance of the Kummer’stest it is mostly ratified by its implications. For instance, Bertrand’s test,Gauss’s test, Raabe’s test [9] are all special cases of Theorem 1. Kummer’stest may be also usefull to characterize convergence in normed vector spaces [6,1.7] and applications of this test can be found in other branches of Analysis,such as difference equations [2], as well.On the other hand, turning our focus to series of the form P c n a n , thereare only few results dealing with type of series. The Abel’s test and test ofDedekind and Du-Bois Reymond (see for instance, [4, p.315], [3] ) are probablythe most famous, since they deal with general series of complex numbers. Thesetests provide conditions that ensure convergence by means of independent as-sumptions on { c n } and { a n } . In this context, the main feature of our results(Theorem 4 and Theorem 5 ) is that they characterizes the relation between thesequences { a n } and { c n } in order to ensure necessary and sufficient conditionsfor the convergence and divergence of the series P c n a n , respectively. Moreover,we present some examples and interesting consequences of this characterization.In particular, generalized versions of Raabe’s, Bertrand’s and Gauss’s test forconvergence and divergence of series of the form P c n a n are obtained. Anotherimportant consequence of Theorem 4 is that it is possible to show that Olivier’stheorem (see, for instance [4, p.124] or [7, 8] for mor information) still holdswhen the monotonicity assumption on the sequence of positive terms { a n } isreplaced by an additional assumption on a auxiliary sequence. We also presentconsequences of Theorem 4 when it is combined to the to the well-know Abelsummation formula and the Cauchy condensation theorem. We refer to [4], p.120 and p. 313 for more details on these results.The rest of the paper is organized as follows. In Section 2, we present nec-essary and sufficient conditions for convergence/divergence of series generatedby subsequences by extending Theorem 1. In Section 3 we present the resultsdealing with convergence and divergence of series of the form P c n a n . The mainideia is to obtain necessary and sufficient conditions by means of an extension ofTheorems 2 and Theorem 3. As we show, we characterize the relation betweenthe sequences { c n } and { a n } that ensures convergence and divergence of theseries. In Section 4 we present some consequences of the results obtained. In this section we present a first extension of Theorem 1. Its main feature isthat it showns that is possible to obtain information about the summability of asequence of positive real numbers based on the relation between non-consecutiveelements of this sequence. In partiular, the idea is to characterize the summa-bility of a sequence by comparing it to the elements of the translated sequence { a n + m , n ≥ } , for some m ≥ Theorem 2.
Let { a n } be a sequence of positive real numbers and m ≥ anyfixed positive integer.If that there exists a positive sequence { q n } such that q n a n a n + m − q n + m ≥ c, or some c > , for all n sufficiently large, then P a n converges. The converseholds as well.Proof. From the assumption we get that q n a n − a n + m q n + m ≥ ca n + m , for all n > N , for some N large. Hence q n a n − q n +1 a n +1 + q n +1 a n +1 − q n +2 a n +2 + ...... − q n + m − a n + m − + q n + m − a n + m − − q n + m a n + m ≥ ca n + m , for all n > N . That is, N + k X n = N +1 q n a n − a n + m q n + m ≥ c N + k X n = N +1 a n + m , for all k ≥
1. That is, by the telescopic sum, q N +1 a N +1 − a N + k + m +1 q N + k + m +1 ≥ c N + k X n = N +1 a n + m , for all k ≥
1. This last inequality implies the convergence of the series P a n + m and therefore, P a n converges.Conversely, if P a n converges, P a n = S say, then let us write write P a n + m = S m , for m ≥
1, positive integer. Let us define { q n } as q n = S m − P ni =1 a i + m a n , n = 1 , , ... , thus, for this { q n } we have that q n a n a n + m − q n + m = P n + mi = n +1 a i + m a n + m = 1 + a n +1 + ... + a n + m − + a n + m +1 + ... + a n +2 m a n + m > , for all n ≥
1. The proof is concluded.We proceed by presenting a divergence version for the previous theorem.
Theorem 3.
Let { a n } be a sequence of positive real and m ≥ a fixed positiveinteger. If there exists a positive sequence { q n } such that P q n diverges, q n a n ≥ c > , and q n a n a n + m − q n + m ≤ , for all n sufficiently large, then P ∞ n =1 a n diverges. The converse holds, as well. roof. From the assumptions we obtain that there exists
N > q n a n a n + m − q n + m ≤ , for all n ≥ N . As so, c q n + m ≤ a n + m , for all n > N . Since P /q n diverges, we obtain from the comparsion test that P a n diverges.Conversely, suppose that P a n diverges. Define for each n ≥ q n = P ni =1 a i a n . Note that the definition implies q = 1, hence a n q n = P ni =1 a i ≥ a , for all n ≥
1, that is, a n q n ≥ a q > n ≥
1. Clearly q n a n a n + m − q n + m ≤ , for all n ≥ P q n diverges. From the divergence of P a n , givenany positive integer k there exists a positive integer n ≥ k such that a k + ... + a n ≥ a + ... + a k − . (1)Due to (1), n X j = k q j = a k a + ... + a k + ... + a n a + ... + a n = 1 a + ... + a k − a k + ... + a n + 1 > . Hence, P nj =1 1 q j is not a Cauchy sequence. Therefore the series P q n diverges. Let us now turn our atention to series of the form P c n a n with positive terms.The central idea in the following result is that it characterizes the relationbetween the sequences { c n } and { a n } in order to ensure the convergence ofthe series. The reader will note that the proof follows the same lines as theproof of Theorem 2 and also, that it could be obtained by some changes inthe proof of Theorem 1, nevertherless, as the reader will also note, our proofprovides important informations about the relation between the sequences { a n } and { c n } . 4 heorem 4. Consider the series P c n a n with { a n } { c n } sequences of positivereal numbers.The series P c n a n converges if and only if that there exist a sequence { q n } of positive real numbers and a positive integer N ≥ for which q n a n a n +1 − q n +1 ≥ c n +1 , n ≥ N. Proof.
Let us show that P c n a n converges. For this, note that the condition q n a n a n +1 − q n +1 ≥ c n +1 , n ≥ N implies that a n q n ≥ a n +1 ( q n +1 + c n +1 ) , n ≥ N. (2)That is, a N q N ≥ r k +1 X i =1 c i a i + a N + k +1 q N + k +1 ≥ r k +1 X i =1 c i a i > , for all integer k ≥
0. This implies the convergence of P c n a n .For the converse, suppose that S := P c n a n and let us define q n = S − P ni =1 c i a i a n , n ≥ N. (3)For this { q n } , clearly q n > n ≥ q n a n a n +1 − q n +1 = c n +1 , n ≥ N. Some remarks:( i ) One can observe that it is, of course, possible to reduce any series to thisform, as any number can be expressed as the product of two other numbers.Success in applying the above theorem will depend on the skill with which theterms are so split up.( ii ) Note that in the first part of Theorem 4, the assumption of positivityof the sequences { a n } and { c n } can be replaced by the following assumptions: { a n } is positive and { c n } is such that P ki =1 c i a i > k sufficiently large.Next, we presente a version of the Kummer’s test for divergent series ofthe form P c n a n . The reader will note that it is more restrictive when it iscompared to Theorem 1-( ii ) however it may be suitable in some cases.5 heorem 5. Consider the series P c n a n with { a n } { c n } sequences of positivereal numbers. ( i ) Suppose that there exist a sequence { q n } and a positive integer N forwhich q n a n a n +1 − q n +1 ≤ − c n +1 , n ≥ N with P q n being a divergent series. Then P a n , P c n , P ( q n − c n ) a n and P q n a n diverge. If, in addition, P c n q n diverges then P c n a n diverges. ( ii ) Suppose that both series P c n a n and P a n diverge. Also, suppose thatfor every m ∈ N there exists r ∈ N such that a m + ... + a r ≥ c m a m + ... + c r a r . Then there exist a sequence { q n } and a positive integer N ≥ such that P q n diverges and q n a n a n +1 − q n +1 ≤ − c n +1 , n ≥ N. Proof.
To prove ( i ) note that { q n } satisfies a n +1 ≥ q n a n q n +1 − c n +1 , n ≥ N and also we have that 0 < q n +1 − c n +1 < q n +1 , n ≥ N. (4)In particular 0 < c n +1 < q n +1 , for n ≥ N and by the comparsion test we seethat P c n diverges.Since, a N +1 ≥ q N a N q N +1 − c N +1 ,a N +2 ≥ q N +1 a N +1 q N +2 − c N +2 ≥ a N q N q N +1 ( q N +2 − c N +2 )( q N +1 − c N +1 ) , and in general, a N + k +1 ≥ a N q N q N +1 ...q N + k ( q N +1 − c N +1 ) ... ( q N + k +1 − c N + k +1 ) , k ≥ , (5)from (4) this last inequality reads as a N + k +1 > a N q N + k +1 , k ≥ . (6)That is, ∞ X k =0 a N + k +1 > a N ∞ X k =0 q N + k +1 , P a n diverges. From (5)( q N + k +1 − c N + k +1 ) a N + k +1 ≥ a N q N q N +1 ...q N + k ( q N +1 − c N +1 ) ... ( q N + k − c N + k ) , k ≥ q N + k +1 a N + k +1 > ( q N + k +1 − c N + k +1 ) a N + k +1 ≥ a N q N > , k ≥ . This last set of inequalities implies that lim n →∞ q N + k +1 a N + k +1 = 0 and lim k →∞ ( q N + k +1 − c N + k +1 ) a N + k +1 = 0, so both series P q n a n and P ( q n − c n ) a n diverge.Note that from (6) we obtain that c N + k +1 a N + k +1 > a N c N + k +1 q N + k +1 , k ≥ . (7)Therefore, if P c n q n diverges, then it is clear that P c n a n diverges.In order to prove ( ii ) define q n = P ni =1 c i a i a n , n ≥ . Clearly, this is a sequence of positive real numbers that satisfies q n a n a n +1 − q n +1 ≤ − c n +1 , n ≥ . Let us show that P qn diverges by concluding that the sequence { s k } , definedas s k = P ki =1 1 q i , for each k ≥
1, is not a Cauchy sequence. Since P c n a n isdivergent, given m ∈ N there exists k ∈ N such that c m a m + ... + c k a k > c a + ... + c m − a m − . (8)Also, from the hypothesis, there exists r > m such that a m + ... + a r ≥ c m a m + ... + c r a r . (9)If k ≤ r , then c m a m + ... + c k a k + ... + c r a r > c m a m + ... + c k a k > c a + ... + c m − a m − and by (8) r X n = m q n = a m c a + ... + c m a m + ... + a r c a + ... + c r a r ≥ a m + ... + a r c a + ... + c r a r ≥ c m a m + ... + c r a r c a + ... + c r a r ≥ c a + ... + c m − a m − c m a m + ... + c r a r + 1 > .
7n the other hand, if k > r then there exists r ≥ r + 1 such that a r +1 + ... + a r ≥ c r +1 a r +1 + ... + c r a r . Again, we can use the same argument to conclude that there exists r ≥ r + 1 a r +1 + ... + a r ≥ c r +1 a r +1 + ... + c r a r . This procedure can be applied a finite number of times in order to obtain r j > n for which a r j − +1 + ... + a r j ≥ c r j − +1 a r j − +1 + ... + c r j a r j . Summing up all this previous inequalities we obtain that a m + ... + a r k ≥ c m a m + ... + c r k a r k . This concludes the proof.
The main goal in this section is to present some of the implications of the mainresults of this paper.The next three theorems are extensions of the Raabe, Bertrand and Gausstest derived from Theorem 4 and Theorem 5. For more information about thesetests we refer to [1, 4] and references therein.Consider the sequences R − n = n a n a n +1 − ( n + 1) − c n +1 and R + n = n a n a n +1 − ( n + 1) + c n +1 , for all positive integer n . Theorem 6 (Raabe’s test) . Let P c n a n be a series of positive terms and sup-pose that lim inf R − n = R and lim sup R − n = R . If ( i ) R ≥ , then P c n a n converges; ( ii ) R ≤ and P c n /n diverges then P c n a n diverges.Proof. ( i ) If R ≥
0, then for all n sufficiently large we have that n a n a n +1 − ( n + 1) − c n +1 ≥ R ≥ , hence Theorem 4, with q n = n for all n ≥
1, implies that the series P c n a n converges.( ii ) If R ≤ n sufficiently large n a n a n +1 − ( n + 1) + c n +1 ≤ R ≤ . Again, we have q n = n for all n ≥
1. So, due to the divergence of P c n /n ,Theorem 5 implies that P c n a n diverges.8 heorem 7 (Bertrand’s test) . Let P c n a n be a series of positive terms. ( i ) If a n a n +1 ≥ n + θ n + c n +1 n ln( n ) , for some sequence { θ n } , such that lim inf θ n ≥ , all n sufficiently large, then P c n a n converges. ( ii ) If a n a n +1 ≤ n + θ n − c n +1 n ln( n ) , for some sequence { θ n } , such that lim sup θ n ≤ all n sufficiently large and lim c n / ( n ln( n )) = 0 then P c n a n diverges.Proof. ( i ) From the assumption we get n ln( n ) a n a n +1 ≥ n ln( n ) + ln( n ) + c n +1 + θ n , for all n sufficiently large. That is, n ln( n ) a n a n +1 − ( n + 1) ln( n + 1) ≥ ( n + 1) ln (cid:18) nn + 1 (cid:19) + θ n + c n +1 , for all n sufficiently large. Since lim inf( n + 1) ln (cid:16) nn +1 (cid:17) θ n ≥ n ln( n ) a n a n +1 − ( n + 1) ln( n + 1) ≥ c n +1 , for all n sufficiently large. Therefore the conclusion follows from Theorem 4.( ii ) It suffices to note that n ln( n ) a n a n +1 − ( n + 1) ln( n + 1) ≤ ( n + 1) ln (cid:18) nn + 1 (cid:19) + θ n − c n +1 , for all n sufficiently large. Hence, due to, lim sup( n + 1) ln (cid:16) nn +1 (cid:17) θ n ≤
0, itfollows that n ln( n ) a n a n +1 − ( n + 1) ln( n + 1) ≤ − c n +1 , for all n sufficiently large. The conclusion follows from Theorem 5 since lim c n / ( n ln( n )) =0. Theorem 8 (Gauss’s test) . Let P c n a n be a series of positive, γ ≥ , { θ n } abounded sequence of real numbers. ( i ) Suppose that there exists a µ ∈ R such that θ n ≥ (1 − µ ) n γ − holds forall n sufficiently large. If a n a n +1 ≥ c n +1 n + µn + θ n n γ , olds for all n sufficiently large, then P c n a n converges. ( ii ) Suppose that there exists a µ ∈ R such that θ n ≤ (1 − µ ) n γ − holds forall n sufficiently large. If P c n /n diverges and a n a n +1 ≤ − c n +1 n + µn + θ n n γ , for all n sufficiently large, then P c n a n diverges.Proof. ( i ) From the assumption we obtain that n a n a n +1 − ( n + 1) ≥ c n +1 + ( µ −
1) + θ n n γ − , for all n sufficiently large. Take N > µ − θ n n γ − ≥ n > N ,we concude that n a n a n +1 − ( n + 1) ≥ c n +1 , for all n > N . Therefore, Theorem 4 applies and implies that P c n a n converges.( ii ) Due to the assumptions on ( ii ), we have that µ − < n a n a n +1 − ( n + 1) ≤ − c n +1 + ( µ −
1) + θ n n γ − ≤ − c n +1 , for all n sufficiently large. The conclusion follows from an applications of The-orem 5.For an illustration of the use of Theorem 2 consider the following example.The following well-know result deals with the case when P a n is such that { a n } is a decreasing sequence. Lemma 1. [4, p. 120](Cauchy’s condensation test ) Let { a n } be a decreasingsequence of positive numbers. Then P a n converges if, and only if, P n a n converges. For a decreasing sequence { a n } of positive real numbers, combining Lemma1 with Theorem 4, we obtain a the following characterization of convergence. Theorem 9. P a n converges if, and only if, there exists a sequence { q n } ofpositive numbers such that q n − q n +1 ≥ a n +1 , for all n sufficiently large.Proof. By Lemma 1 P a n coverges if, and only if, P n a n converges. If wedefine a n = 2 n and c n = a n +1 then, by Theorem 4, this last series converges if,and only if, there exists a sequence { q n } of positive real numbers such that q n − q n +1 ≥ a n +1 , for all n sufficiently large. The proof is concluded.10 xample 1. Let us recall the definition of ℓ in order to present an applicationof our previous results. Since, by definition, ℓ = n { a n } ⊂ R , X a n , converges o , an immediate consequence of Theorem 4 is { a n } ∈ ℓ if, and only if, there existsa sequence { q n } of positive numbers such that q n a n − q n +1 a n +1 ≥ a n +1 , for all n sufficiently large. Let { a n } be summable decreasing sequence of positive real numbers. It iswell know that lim n a n = 0. This result is often called Olivier’s Theorem (seepage 124 of [4, p.124] or [7]). We close the paper by showing that it possible torecover the same asymptotic behaviour for { a n } without the decreasigness as-sumption on { a n } . Instead of using the monotonicity, we consider an additionalassumption on the sequence { q n } (that auxiliary sequence of Theorem 4). Letus first introduce an auxiliary notation.We will write { q n } ∈ R + [ { a n }{ c n } ] when { q n } is a sequence of positivenumbers such that q n a n n n +1 − q n +1 ≥ c n +1 , for all n sufficiently large. By Theorem 4 the inequality above is equivalent tothe summability of { a n c n } , when these are sequences of positive numbers. Theorem 10.
Suppose that { a n } is a sequence of positive real numbers suchthat P a n converges. If there exists { q n } ∈ R + [ { na n }{ n } ] such that lim q n n + 1 n − q n +1 = 0 then lim na n = 0 .Proof. Let { a n } be as stated. If P a n converges then it is clear that P n na n also converges. From Theorem 4, with a n = 1 /n and c n = na n , we can concludethat P a n converges if, and only if, there exists a sequence { q n } such that q n n + 1 n − q n +1 ≥ ( n + 1) a n +1 , for all n sufficiently large. Hence, lim na n = 0 certainly occurs when the se-quence { q n } above is such thatlim q n n + 1 n − q n +1 = 0 . For more information on this asymptotic behavior of summable sequencesof positive numbers we refer to [7, 5, 8] and references therein.11 eferences [1] L. Bourchtein; A. Bourchtein; G. Nornberg and C. Venzke, A Hierarchyof the Convergence Tests Related to Cauchy’s Test. Int. Journal of Math.Analysis, , ( 2012), 1847–1869.[2] I. Gyori,; L. Horvath, l p -solutions and stability analysis of difference equa-tions using the Kummer’s test, Applied Mathematics and Computation, , (2011), 10129–10145.[3] J. Hadamard, Deux theoremes d’Abel sur la convergence des series, Acta.Math., , (1903), 177–183.[4] K. Knopp, Theory and application of infinite series, Blackie & Son Limited ,London and Glascow, 1954.[5] E. Liflyanda, S. Tikhonovb, M. Zeltserc, Extending tests for convergence ofnumber series. J. Math. Anal. Appl., , (2011), 194—06.[6] J. Muscat, Functional Analysis
An Introduction to Metric Spaces, HilbertSpaces, and Banach Algebras,
Springer, 2014.[7] C. P. Niculescu; F. Popovici, A note on the behavior of integrable functionsat infinity. J. Math. Anal. Appl., , (2011), 742–747 .[8] T. S´alat, V. Tomma, A Classical Olivier’s Theorem and Statistical Conver-gence, Annales math´ematiques Blaise Pascal, , (2003), 305–313[9] J. Tong, Kummer’s, Test Gives Characterizations for Convergence or Diver-gence of all Positive Series, The American Mathematical Monthly ,5