Infinite Series: The Abel-Dini Theorem and Convergence Tests Logan Gnanapragasam June 10, 2019 1 Introduction In this paper, we will discuss infinite series. In K. Knopp's book [3], a proof that there is no perfect test for convergence is given. To do this, Knopp uses the Abel-Dini Theorem, which is of interest in its own right. The Abel-Dini Theorem is discussed more fully in T. H. Hildebrandt's article [2]. In section 2, we provide a proof of the Abel-Dini Theorem and discuss some applications. In section 3, we will discuss the results in [3] about convergence tests. Fairly little advanced machinery is required to prove the results on convergence tests. Since this is a topic in real analysis, the reader should expect to see many inequalities. They should also be familiar with sequences of numbers. Familiarity with sequences of functions and uniform convergence is recommended but by no means required for the discussion of convergence tests. We will occasionally use lim sup and lim inf, but the reader does not need to know all of the equivalent definitions. We will give the lim sup definition that is most applicable to us below. 1.1 Real Analysis Background Here we give the definitions and basic theorems on series in real analysis that we will need. We do not prove any of these results; proofs can be found in most introductory real analysis books or (advanced) calculus books. Before we discuss series, we define lim sup and lim inf. 1 Definition 1 (lim sup and lim inf). Given a sequence fangn=1, we say that M = lim sup an n!1 if • for each u > M there are only finitely many indices n for which an > u, and • for each l < M there are infinitely many indices n for which an > l. Similarly, we say m = lim inf an if n!1 • for each l < m there are only finitely many indices n for which an < l, and 1 • for each u > m then there are infinitely many indices n for which an < u. We allow lim sup and lim inf to be ±∞, removing the appropriate condition in the standard definition. For example, we say lim sup an = 1 if for each l 2 R there are infinitely many n!1 indices n for which an > l. The remaining modifications are left to the reader. Every sequence has a lim sup and lim inf, and this definition uniquely specifies these values; see Exercise 1.5.9 and the discussion preceeding it in [1]. We will not prove these facts. Now we turn to infinite series. We begin with series of numbers. 1 1 X X Definition 2 (Convergence of a Series of Numbers). A series an, often written an n=1 1 1 when the index of summation is unambiguous, converges if the sequence fSngn=1 defined n X by Sn = ak converges (as a sequence of numbers). If Sn ! S, we say that the sum of k=1 the series is S. 1 X If the starting index of summation is j, so that the series is an, we modify the n=j N X definition by declaring SN = an. If the starting index is unknown or irrelevant, we will n=j 1 X X write an instead of an. It is useful to give a name to the sequence associated with j the series. 1 N X X Definition 3 (Partial Sums). Given a series an, we say that SN = an is the Nth n=1 1 partial sum of the series. Thus our definition of convergence is that a series converges if the sequence of partial sums converges. 1 1 X X Observe that whenever the series an converges, so does an, since the constant n=1 n=N N−1 X an has been subtracted off from every partial sum of the original series. This observation n=1 justifies the following definition. 1 1 X X Definition 4 (Tails). If an converges, we say that an is the Nth tail of the series. n=1 n=N The indexing in this definition differs from the definition given in some other texts. Some 1 X authors call an the Nth tail. The difference is not significant as long as the theorems n=N+1 are stated correctly, but the reader should be aware of this distinction. 2 The following result follows immediately from the fact that the sequence of partial sums converges if and only if it is Cauchy. 1 X Theorem 1 (Cauchy Convergence Criterion). The series ak converges if and only if for 1 m X each > 0, there is N such that ak < whenever m ≥ n ≥ N. k=n From this, one can prove the basic comparison test for series of nonnegative terms. Also, by applying the triangle inequality, one finds that absolute convergence implies convergence. For the statements of these results, see Theorems 6.11 and 6.17 in [1]. On occasion we will have something to say about uniform convergence of a series of functions. The definition of uniform convergence of a sequence of functions can be found in Section 7.1 of [1]. Definition 5 (Uniform Convergence of a Series of Functions). Given a sequence of real- 1 1 X valued functions ffngn=1 defined on a set S, we say that the series of functions fn n=1 N X converges uniformly on a set W ⊆ S if the sequence of partial sums SN = fn, which n=1 are also real-valued functions defined on S, converges uniformly on W . We also have an analogous result to the Cauchy Convergence Criterion, which follows from the fact that a sequence of functions converges uniformly if and only if the sequence is uniformly Cauchy (Theorem 7.7 in [1]). Theorem 2 (Cauchy Convergence Criterion for Functions). Given a sequence of functions 1 X fk : S ! R, fk converges uniformly on S if and only if for each > 0, there is N such 1 m X that fk(x) < for all x 2 S whenever m ≥ n ≥ N. k=n It is possible to prove the Weierstrass M-Test (Theorem 7.9 in [1]) with the triangle in- equality and the Cauchy Convergence Criterion for numbers and the Cauchy Convergence Criterion for functions. While the Cauchy Convergence Criterion is important, it is not always easy to apply. There are several tests that can be used to determine whether a series converges. The first is the standard limit comparison test, not stated here (see Theorem 6.12 in [1]). This isn't always easy to apply because we need to come up with a series to compare to. Two of the best known convergence tests which do not require us to find a comparison series are the ratio and root tests. 1 X ak+1 Theorem 3 (Ratio Test). If a is a series of positive terms, u = lim sup , and k a 1 k!1 k a l = lim inf k+1 , then the series converges if u < 1 and diverges if l > 1. Otherwise no k!1 ak conclusions can be drawn. 3 1 X 1 Theorem 4 (Root Test). If ak is a series of nonnegative terms and r = lim sup (ak) k , 1 k!1 then the series converges if r < 1 and diverges if r > 1. If r = 1 then no conclusions can be drawn. It is possible to prove that if all the an are positive, then 1 1 ak+1 ak+1 lim inf ≤ lim inf (ak) k ≤ lim sup (ak) k ≤ lim sup ; k!1 ak k!1 k!1 k!1 ak which shows that if the ratio test gives the convergence/divergence of a particular series then so does the root test (i.e., the root test is better than the ratio test). However, the 1 X root test can also be inconclusive; both the divergent series 1 and the convergent series 1 1 X 1 are series on which the root test gives r = 1. To understand why r = 1 is an issue, n2 1 we need to examine the proof of the root test. For convergence when r < 1, the series is X compared to a geometric series xn with r < x < 1. For divergence when r > 1, one observes that janj ≥ 1 infinitely often, so an 6! 0 and hence the series diverges. However, since the geometric series has radius of convergence 1, when r = 1 we can't compare to a geometric series, but we also cannot say that the terms are bigger than 1 in absolute value infinitely often. Inconclusiveness is not specific to tests relying on comparison to geometric series. As an example, we briefly discuss Raabe's Test. 1 X an+1 Theorem 5 (Raabe's Test). If a is a series of positive terms for which ! 1 and k a 1 n a n 1 − n+1 ! L as n ! 1, then the series converges if L > 1 and diverges if L < 1. If an L = 1 then no conclusions can be drawn. For a more complete discussion of Raabe's Test, along with a proof, see the subsection \Raabe's Test" in Section 6.2 of [1]. The proof involves comparing the series to the p-series for a well-chosen p. As we will see in Section 2.1, the p-series converges if and only if p > 1. When L = 1, we cannot compare to the p-series and hence the test is inconclusive. Notice that all of our tests rely on comparison.