8.1. NUMERICAL SERIES AND INFINITE PRODUCTS 339 If D < 1, then the series ∞ n=1 a n is convergent. If D > 1, then the series is divergent. For D = 1, the D’Alembert criterion cannot be used for deciding whether the series is convergent or divergent. Example 1. Let us examine convergence of the series ∞ n=1 n k x n with x > 0, using the D’Alembert criterion. Taking a n = n k x n ,weget a n+1 a n = 1 + 1 n k x → x as n →∞. Therefore, D = x. It follows that the series is convergent for x < 1 and divergent for x > 1. 4. Cauchy’s criterion. Suppose that there exists the limit (finite or infinite) lim n→∞ n √ a n = K. For K < 1,theseries ∞ n=1 a n is convergent; for K > 1, the series is divergent. For K = 1, the Cauchy criterion cannot be used to establish convergence of a series. Remark. The Cauchy criterion is stronger than the D’Alembert criterion, but the latter is simpler than the former. 5. Gauss’ criterion. Suppose that the ratio of two consecutive terms of a series can be represented in the form a n a n+1 = λ + μ n + o 1 n as n →∞. The series ∞ n=1 a n is convergent if λ > 1 or λ = 1, μ > 1. The series is divergent if λ < 1 or λ = 1, μ ≤ 1. 6. Maclaurin–Cauchy integral criterion.Letf(x) be a nonnegative nonincreasing continuous function on the interval 1 ≤ x < ∞.Letf(1)=a 1 , f(2)=a 2 , , f(n)=a n , Then the series ∞ n=1 a n is convergent if and only if the improper integral ∞ 1 f(x) dx is convergent. Example 2. The harmonic series ∞ n=1 1 n = 1 + 1 2 + 1 3 + ··· is divergent, since the integral ∞ 1 1 x dx is divergent. In a similar way, one finds that the series ∞ n=1 1 n α is convergent for α > 1 and divergent for α ≤ 1. 8.1.2-2. Other criteria of convergence (divergence) of series with positive terms. 1. Raabe criterion. Suppose that there exists the limit (finite or infi nite) lim n→∞ n a n a n+1 – 1 = R. Then, for R > 1 the series ∞ n=1 a n is convergent, and for R < 1 it is divergent. For R = 1, the Raabe criterion is inapplicable. 340 SERIES 2. Bertrand criterion. Suppose that there exists the limit (finite or infi nite) lim n→∞ ln n n a n a n+1 – 1 – 1 = B. Then, for B > 1 the series ∞ n=1 a n is convergent, and for B < 1 it is divergent. For R = 1,the Bertrand criterion is inapplicable. 3. Kummer criterion.Letb n be an arbitrary sequence with positive terms. Suppose that there exists the limit (finite or infinite) lim n→∞ b n a n a n+1 – b n+1 = K. Then, for K > 0,theseries ∞ n=1 a n is convergent. If K < 0 and the additional condition ∞ n=1 1 b n = ∞ holds, then the series is divergent. Remark. From the Kummer criterion, we obtain the D’Alembert criterion (taking b n = 1), the Raabe criterion (taking b n = n), and the Bertrand criterion (taking b n = n ln n). 4. Ermakov criterion.Letf (x) be a positive monotonically decreasing continuous function on the interval 1 ≤ x < ∞.Letf (1)=a 1 , f(2)=a 2 , , f (n)=a n , Then the following implications hold: 1)If e x f(e x ) f(x) ≤ q < 1, then the series ∞ n=1 a n is convergent. 2)If e x f(e x ) f(x) ≥ 1, then the series ∞ n=1 a n is divergent. Here, it suffices to have the inequalities on the left for sufficiently large x ≥ x 0 . 5. Generalized Ermakov criterion.Letf(x) be the function involved in the Ermakov criterion, and let ϕ(x) be an arbitrary positive monotonically increasing function that has a continuous derivative and satisfies the inequality ϕ(x)>x. Then the following implications hold: 1)If ϕ(x)f ϕ(x) f(x) ≤ q < 1, then the series ∞ n=1 a n is convergent. 2)If ϕ(x)f ϕ(x) f(x) ≥ 1, then the series ∞ n=1 a n is divergent. Here, it suffices to have the inequalities on the left for sufficiently large x ≥ x 0 . 6. Sapogov criterion.Leta 1 , a 2 , be a monotonically increasing sequence. Then the series ∞ n=1 1 – a n a n+1 as well as ∞ n=1 a n a n+1 – 1 is convergent, provided that the sequence a n is bounded (a n ≤ L). Otherwise, this series is divergent. 8.1. NUMERICAL SERIES AND INFINITE PRODUCTS 341 7. Special Cauchy criterion. Suppose that a 1 , a 2 , is a monotonically decreasing sequence. Then the series ∞ n=1 a n is convergent (resp., divergent) if and only if the series ∞ n=1 2 n a 2 n is convergent (resp., divergent). 8. Abel–Dini criterion.Iftheseries ∞ n=1 a n is divergent and s n denotes its partial sum, then the series ∞ n=1 a n s n is also divergent, while the series ∞ n=1 a n s 1+σ n (σ > 0)isconvergent. 9. Dini criterion.Iftheseries ∞ n=1 a n is convergent and γ n denotes its remainder after the nth term, then the series ∞ n=1 a n γ n–1 is divergent, while the series ∞ n=1 a n γ σ n–1 (0 < σ < 1)is convergent. 10. Bugaev criterion. If the function ϕ (x)u ϕ(x) is monotone for large enough x, then the series ∞ n=1 u(n)and ∞ n=1 ϕ (n)u ϕ(n) are convergent or divergent simultaneously. 11. Lobachevsky criterion.Letu(x) be a monotonically decreasing function defined for all x. Then the series ∞ n=1 u(n) is convergent or divergent if and only if the series ∞ k=1 p k 2 –k is convergent or divergent, where p k is defined from the relation u(p k )=2 –k . 8.1.3. Convergence Criteria for Arbitrary Numerical Series. Absolute and Conditional Convergence 8.1.3-1. Arbitrary series. Leibnitz, Abel, and Dirichlet convergence criteria. 1. Leibnitz criterion. Suppose that the terms a n of a series ∞ n=1 a n have alternating signs, their absolute values form a nonincreasing sequence, and a n → 0 as n →∞. Then this “alternating” series is convergent. If S is the sum of the series and s n is its nth partial sum, then the following inequality holds for the error |S – s n | ≤ |a n+1 |. Example 1. The series 1 – 1 2 2 + 1 3 3 – 1 4 4 + 1 5 5 – ··· is convergent by the Leibnitz criterion. Taking S ≈ s 4 = 1 – 1 2 2 + 1 3 3 – 1 4 4 , we obtain the error less than a 5 = 1 5 5 = 0.00032. 2. Abel criterion. Consider the series ∞ n=1 a n b n = a 1 b 1 + a 2 b 2 + ···+ a n b n + ···,(8.1.3.1) where a n and b n are two sequences or real numbers. Series (8.1.3.1) is convergent if the series ∞ n=1 b n = b 1 + b 2 + ···+ b n + ··· (8.1.3.2) is convergent and the a n form a bounded monotone sequence (|a n | < K). 342 SERIES 3. Dirichlet criterion. Series (8.1.3.1) is convergent if partial sums of series (8.1.3.2) are bounded uniformly in n, n k=1 b k ≤ M (n = 1, 2, ), and the sequence a n → 0 is monotone. Example 2. Consider the series ∞ n=1 a n sin(nx), where a n → 0 is a monotonically decreasing sequence. Taking b n =sin(nx) and using a well-known identity, we find the partial sum s n = n k=1 sin(kx)= cos 1 2 x –cos n + 1 2 x 2 sin 1 2 x (x ≠ 2mπ; m = 0, 1, 2, ). This sum is bounded for x ≠ 2mπ: |s n | ≤ 1 sin 1 2 x . Therefore, by the Dirichlet criterion, the series ∞ n=1 a n sin(nx) is convergent for any x ≠ 2mπ. Direct verification shows that this series is also convergent for x = 2mπ (since all its terms at these points are equal to zero). Remark. The Leibnitz and the Able criteria can be deduced from the Dirichlet criterion. 8.1.3-2. Absolute and conditional convergence. 1. Absolutely convergent series.Aseries ∞ n=1 a n (with terms of arbitrary sign) is called absolutely convergent if the series ∞ n=1 |a n | is convergent. Any absolutely convergent series is convergent. In order to establish absolute conver- gence of a series, one can use all convergence criteria for series with nonnegative terms given in Subsection 8.1.2 (in these criteria, a n should be replaced by |a n |). Example 3. The series 1 + 1 2 2 – 1 3 2 – 1 4 2 + 1 5 2 + 1 6 2 – ··· is absolutely convergent, since the series with the absolute values of its terms, ∞ n=1 1 n 2 , is convergent (see the second series in Example 2 of Subsection 8.1.2 for α = 2). 2. Conditionally convergent series. A convergent series ∞ n=1 a n is called conditionally convergent if the series ∞ n=1 |a n | is divergent. Example 4. The series 1 – 1 2 + 1 3 – 1 4 + ··· is conditionally convergent, since it is convergent (by the Leibnitz criterion), but the series with absolute values of its terms is divergent (it is a harmonic series; see Example 2 in Subsection 8.1.2). Any rearrangement of the terms of an absolutely convergent series (in particular, a convergent series with nonnegative terms) neither violates its absolute convergence nor changes its sum. Conditionally convergent series do not possess this property: the terms of a conditionally convergent series can be rearranged in such order that the sum of the new series becomes equal to any given value; its terms can also be rearranged so as to result in a divergent series. 8.1. NUMERICAL SERIES AND INFINITE PRODUCTS 343 8.1.4. Multiplication of Series. Some Inequalities 8.1.4-1. Multiplication of series. Cauchy, Mertens, and Abel theorems. A product of two infinite series ∞ n=0 a n and ∞ n=0 b n is understood as a series whose terms have the form a n b m (n, m = 0, 1, ). The products a n b m can be ordered to form a series in many different ways. The following theorems allow us to decide whether it is possible to multiply series. C AUCHY THEOREM. Suppose that the series ∞ n=0 a n and ∞ n=0 b n are absolutely convergent and their sums are equal to A and B , respectively. Then any product of these series is an absolutely convergent series and its sum is equal to AB . The following Cauchy multiplication formula holds : ∞ n=0 a n ∞ n=0 b n = ∞ n=0 n m=0 a m b n–m .(8.1.4.1) M ERTENS THEOREM. The Cauchy multiplication formula (8.1.4.1) is also valid if one of the series, ∞ n=0 a n or ∞ n=0 b n , is absolutely convergent and the other is (conditionally) conver- gent. In this case, the product is a convergent series, possibly, not absolutely convergent. ABEL THEOREM. Consider two convergent series with sums A and B . Suppose that the product of these series in the form of Cauchy (8.1.4.1) is a convergent series with sum C . Then C = AB . 8.1.4-2. Inequalities. 1. Generalized triangle inequality: ∞ n=1 a n ≤ ∞ n=1 |a n |. 2. Cauchy inequality (Cauchy–Bunyakovsky inequality): ∞ n=1 a n b n 2 ≤ ∞ n=1 a 2 n ∞ n=1 b 2 n . 3. Minkowski inequality: ∞ n=1 |a n + b n | p 1 p ≤ ∞ n=1 |a n | p 1 p + ∞ n=1 |b n | p 1 p , p ≥ 1. 4. H ¨ older inequality (for p = 2 coincides with the Cauchy inequality): ∞ n=1 a n b n ≤ ∞ n=1 |a n | p 1 p ∞ n=1 |b n | p p–1 p–1 p , p > 1. 5. An inequality with π: ∞ n=1 a n 4 ≤ π 2 ∞ n=1 a 2 n ∞ n=1 n 2 a 2 n . In all these inequalities it is assumed that the series in the right-hand sides are convergent. 344 SERIES 8.1.5. Summation Methods. Convergence Acceleration 8.1.5-1. Some simple methods for calculating the sum of a series. THEOREM 1. Suppose that the terms of a series ∞ n=1 a n can be represented in the form a n = b n – b n+1 ,where b n is a sequence with a finite limit b ∞ .Then ∞ n=1 a n = b 1 – b ∞ . Example 1. Let us find the sum of the infinite series S = ∞ n=1 1 n(n + 1) . We have a n = 1 n(n + a) = 1 n – 1 n + 1 =⇒ b n = 1 n . Since b 1 = 1 and lim n→∞ b n = 0,wegetS = 1. THEOREM 2. Suppose that the terms of a series ∞ n=0 a n can be represented in the form a n = b n – b n+m , where m is a positive integer and the sequence b n has a finite limit b ∞ .Then ∞ n=1 a n = b 0 + b 1 + ···+ b m–1 – mb ∞ . T HEOREM 3. Suppose that the terms of a series ∞ n=0 a n can be represented in the form a n = α 1 b n+1 + α 2 b n+2 + ···+ α m b n+m ,(8.1.5.1) where m ≥ 2 is a fixed positive integer, the sequence b n has a finite limit b ∞ ,and α k satisfy the condition α 1 + α 2 + ···+ α m = 0.(8.1.5.2) Then the series is convergent and ∞ n=0 a n = α 1 b 1 +(α 1 +α 2 )b 2 +···+(α 1 +α 2 +···+α m–1 )b m–1 +[α 2 +2α 3 +···+(m–1)α m ]b ∞ . (8.1.5.3) Example 2. For the series ∞ n=0 4n + 6 (n + 1)(n + 2)(n + 3) ,wehave a n = 4n + 6 (n + 1)(n + 2)(n + 3) = 1 n + 1 + 2 n + 2 – 3 n + 3 , which corresponds to the following values in (8.1.5.1): α 1 = 1, α 2 = 2, α 3 =–3, b n = 1 n . Thus, condition (8.1.5.2) holds: 1 + 2 – 3 = 0, and the sequence b n tends to zero: b ∞ = 0. Using (8.1.5.3), we find the sum of the series ∞ n=0 4n + 6 (n + 1)(n + 2)(n + 3) = 1 ⋅ 1 +(1 + 2) 1 2 = 5 2 . 8.1. NUMERICAL SERIES AND INFINITE PRODUCTS 345 8.1.5-2. Summation of series with the help of Laplace transforms. Let a(k) be the Laplace transform of a given function f(x), a(k)= ∞ 0 e –kx f(x) dx. Then the following summation formulas hold: ∞ k=1 a(k)= ∞ 0 f(x) dx e x – 1 , ∞ k=1 (–1) k a(k)=– ∞ 0 f(x) dx e x + 1 , (8.1.5.4) provided that the series are convergent. Example 3. It is easy to check that a k 2 + a 2 = ∞ 0 e –kx sin(ax) dx. Therefore, using the first formula in (8.1.5.4), we get ∞ k=1 1 k 2 + a 2 = 1 a ∞ 0 sin(ax) dx e x – 1 = π 2a coth(πa)– 1 2a 2 . 8.1.5-3. Kummer and Abel transformations. Acceleration of convergence of series. 1 ◦ . Kummer transformation. Consider a series with positive (nonnegative) terms ∞ n=1 a n (8.1.5.5) and an auxiliary series with a finite sum B = ∞ n=1 b n .(8.1.5.6) Suppose that there is a finite limit K = lim n→∞ b n a n ≠ 0. Under these conditions, the series (8.1.5.5) is convergent and the following identity holds: ∞ n=1 a n = B K + ∞ n=1 1 – 1 K b n a n a n .(8.1.5.7) The right-hand side of (8.1.5.7) is called the Kummer transformation of the given series. . 345 8.1.5-2. Summation of series with the help of Laplace transforms. Let a(k) be the Laplace transform of a given function f(x), a(k)= ∞ 0 e –kx f(x) dx. Then the following summation formulas hold: ∞ k=1 a(k)= ∞ 0 f(x). one finds that the series ∞ n=1 1 n α is convergent for α > 1 and divergent for α ≤ 1. 8.1.2-2. Other criteria of convergence (divergence) of series with positive terms. 1. Raabe criterion nite) lim n→∞ n a n a n+1 – 1 = R. Then, for R > 1 the series ∞ n=1 a n is convergent, and for R < 1 it is divergent. For R = 1, the Raabe criterion is inapplicable. 340 SERIES 2. Bertrand criterion. Suppose