By Krantz S.G.
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2. ELEMENTARY CONVERGENCE TESTS It follows that the series fore it converges. 49 bj satisfies the Cauchy criterion for series. 16 If j=1 aj is as in the proposition and if 0 ≤ bj ≤ aj for ∞ every j then the series j=1 bj converges. Proof: Obvious. 17 The series ∞ with the series j=1 2−j . 18 (The Cauchy Condensation Test) Assume that a1 ≥ a2 ≥ · · · ≥ aj ≥ . . 0. The series ∞ aj j=1 converges if and only if the series ∞ k=1 2 k · a2 k converges. Proof: First assume that the series k ≥ 1, 2k−1 · a2k = ∞ j=1 aj converges.
Otherwise, by the Cauchy test, it diverges. 33, we learn the Integral Test. This gives another nice way to think about this example. 24 Let us apply the Cauchy Condensation Test to the series ∞ j=1 1 . j(log2 j)2 This leads us to examine the series ∞ k=1 2k · 1 = 2k · k 2 ∞ k=1 1 . k2 By the preceding corollary, this is a convergent series. So the original series converges. 25 (The Root Test for Convergence) Consider the series ∞ aj . j=1 If 1/j lim sup |aj | j→∞ then the series converges. 2. ELEMENTARY CONVERGENCE TESTS 53 Proof: Refer again to the discussion of the concept of limit superior in Chapter 2.
16 If j=1 aj is as in the proposition and if 0 ≤ bj ≤ aj for ∞ every j then the series j=1 bj converges. Proof: Obvious. 17 The series ∞ with the series j=1 2−j . 18 (The Cauchy Condensation Test) Assume that a1 ≥ a2 ≥ · · · ≥ aj ≥ . . 0. The series ∞ aj j=1 converges if and only if the series ∞ k=1 2 k · a2 k converges. Proof: First assume that the series k ≥ 1, 2k−1 · a2k = ∞ j=1 aj converges. Notice that, for each a2 k + a2 k + · · · + a2 k 2k−1 times ≤ a2k−1 +1 + a2k−1 +2 + · · · + a2k . 2k = am m=2k−1 +1 Therefore N k=1 N 2k−1 · a2k ≤ 2k 2N am = k=1 m=2k−1 +1 am .