CONVERGENCE TESTS 433 15.3 CONVERGENCE TESTS There exist a number of tests for checking the convergence of a given series. In what follows we give some of the most commonly used tests for convergence. The tests are ordered in increasing level of complexity. In practice one starts with the simplest test and, if the test fails, moves on to the next one. In the following tests we either consider series with positive terms or take the absolute value of the terms; hence we check for absolute convergence. 15.3.1 Comparison Test The simplest test for convergence is the comparison test. We compare a given series term by term with another series convergence or divergence of which has been established. Let two series with the general terms a, and b, be given. For all n 2 1 if la,l 5 Ib,l is true and if the series c,"==, lbnl is convergent, then the series Cr=p=I a, is also convergent. Similarly, if xr=, a, is divergent, then the series C,"=l lbnl is also divergent. Example 15.3. Comparison test: Consider the series with the general term a, = n-P where p = 0.999, We compare this series with the harmonic series which has the general term b, = n-'. Since for n 2 1 we can write n-' < n-0.999 and since the harmonic series is divergent, we also conclude that the series xr=l n-p is divergent. 15.3.2 Ratio Test For the series c,"==, a,, let a, # 0 for all n 2 1. When we find the limit an+ 1 lim - n-wl a, (15.8) for r < 1 the series is convergent, for T > 1 the series is divergent, and for r = 1 the test is inconclusive. 15.3.3 Cauchy Root Test For the series x,"=l a,, when we find the limit lim = 1, n-w (15.9) for 1 < 1 the series is convergent, for 1 > 1 the series is divergent, and for 1 = 1 the test is inconclusive. 434 INFINITE SERIES -< I f(4) f(5) ; Ill1 ! IIII IIII I IIII - 12345 X fig. 15.1 Integral test 15.3.4 Integral Test Let a, = f (n) be the general term of a given series with positive terms. If for n > 1, f (n) is continuous and a monotonic decreasing function, that is, f (n+ 1) < f (n), then the series converges or diverges with the integral J;” f (x) fh. Pro0 f As shown in Figure 15.1, we can put a lower and an upper bound to the series C,”==, a, as (15.10) (15.11) From here it is apparent that in the limit as N -+ co, if the integral J: f (x) dx is finite, then the series CT=l an is convergent. If the inte- gral diverges, then the series also diverges. Example 15.4. Integral test: Let us consider the Riemann zeta function 11 2” 35 <(s)= I+-+-+. (15.12) To use the ratio test we make use of the binomial formula and write S I +- n (15.13) CONVERGENCE TESTS 435 f 1; thus the ratio test fails. an+ 1 In the limit as n + this gives - an However, using the integral test we find 1 s > 1 + series is convergent 0;) s < 1 + series is divergent (15.14) 15.3.5 Raabe Test For a series with positive terms, a, > 0, when we find the limit lim n (L - 1) = m, (15.15) for m > 1 the series is convergent and for m < 1 the series is divergent. For m = 1 the Raabe test is inconclusive. The Raabe test can also be expressed as follows: Let N be a positive integer independent of n. For all n 2 N, if n (e - 1) 2 P > 1 is true, then the series is convergent and if n (e - 1) 5 1 is true, then the series is divergent. Example 15.5. Raabe test: For the series c,"==, 5 the ratio test is incon- nioo an+] clusive. However, using the Raabe test we see that it converges: lim n (5 - 1) = lim n (% n+ 112 - 1) (15.16) n-m ,+03 an+] n-cc Example 15.6. Raabe test: The second form of the Raabe test shows that is divergent. This follows from the fact that the harmonic series C,"xl for all n values, (15.17) When the available tests fail, we can also use theorems like the Cauehy theorem. 15.3.6 Cauchy Theorem A given series C,"==, a, with positive decreasing terms (a, 2 a,+l 2 . . . 2 0) converges or diverges with the series 00 (c an integer). (15.18) 2 c cnacn = ca, + c a,2 + c3ac3 + . . . n= 1 436 INFINITE SERIES Example 15.7. Cuwhy theorem; Let us check the convergence of the se- ries 1 nlna n' 03 1 1 fT+.'.= 1 c- +- n= 2 21na2 3ln"3 4111 4 (15.19) by using the Cauchy theorem for (Y 2 0. Choosing the value of c as two, we construct the series xr=l 2"~" = 2a2 + 4a4 + 8as +. . . , where the general term is given as (15.20) Since the series C,"=* -& converges for a > 1, our series is also convergent for (Y > 1. On the other hand, for (Y 5 1, both series are divergent. 15.3.7 Legendre series are given as Gauss Test and Legendre Series Cn=Oa2nz2n 00 and C~=oa2n+lz2n+1, 3: E [-I, 11. (15.21) Both series have the same recursion relation , n=O,1, . (n - I) (I + n + 1) (n + 1) (n + 2) an+2 = an (15.22) For 121 < 1, convergence of both series can be established by using the ratio test. For the even series the general term is given as un = a2nz2n; hence we write (an - I) (an + I + 1) z2 Un QnX2n (an + 1) (an + 2) ' un+1 - a2n+1x2n+1 - - (15.23) (15.24) Using the ratio test we conclude that the Legendre series with the even terms is convergent for the interval z E (-1,l). The argument and the conclusion for the other series are exactly the same. However, at the end points the ratio test fails. For these points we can use the Gauss test: Gauss test: Let C,"==, u, be a series with positive terms. If for n 2 N (N is a given constant) we can write (15.25) CONVERGENCE TESTS 437 where 0 (5) means that for a given function f (n) thelimit limn+OO{.f (n) /$} is finite, then the C,"==, un series converges for p > 1 and diverges for p 5 1. Note that there is no case here where the test fails. Example 15.8. Legendre series: We now investigate the convergence of the Legendre series at the end points, z = fl, by using the Gauss test. We find the required ratio as - (15.26) 21, - (2n+1)(2n+2) un+l 4n2 + 6n + 2 4n2 + 2n - 1 (I + I)' - - - (2n - 1) (272 + 1 + 1) (15.27) Un 1 1(1+ 1) (1 fn) -N [4n2 + 2n - I(1f l)]n' -I+;+ Un+1 From the limit (15.28) I(l+l)(l+n) 1 1(l+l) lim n+m [4n2 + 2n - 1 (I + l)] n"2) = 4 ' we see that this ratio is constant and goes as O(3). Since p = 1 in -, we conclude that the Legendre series (both the even and the odd series) diverge at the end points. Un un+ 1 Example 15.9. Chebyshev series: The Chebyshev equation is given as d2Y dY (1 - 2)- - z- + n2y = 0. dx2 dx (15.29) Let us find finite solutions of this equation in the interval z E [-1,1] by using the Frobenius method. We substitute the following series and its derivatives into the Chebyshev equation: W (15.30) (15.31) k=O 00 y" = ak(k f C?)(k f (Y - 1)xk++"p2 (15.32) k=O to get 438 /NF/N/TE SERIES After rearranging we first get 03 (a - 1) ZQ-2 f a](Y (a f l)xa-' f ak(k f a)(k f a - 1)Xk+a-2 k=2 00 f akxk+a [n2 - (k f O)'] = 0 (15.34) k-0 and then 03 UOQ. (Q. - 1) Za-2 + a]a (a + 1) Zap' f ak+2(k f (Y f 2)(k f a f I)%"+" k=O 03 + akZk+a [n2 - (k f a)2] = 0. (15.35) k=O This gives the indicial equation as aoa (a - 1) = 0, a0 # 0. (15.36) The remaining coefficients are given by a1a (a + 1) = 0 (15.37) and the recursion relation (15.38) Since a0 # 0, roots of the indicial equation are 0 and 1. Choosing the smaller root gives the general solution with the recursion relation Ic2 - n2 (k f 2)(k + qUk ak+2 = (15.39) and the series solution of the Chebyshev equation is obtained as n2 n2 (22 -n2 4-3-2 (1 - n2)Z3 + (32 - n2) (1 - n2) 5.4.3.2 +a1 Xf- ( 3.2 We now investigate the convergence of these series. Since the argument for both series is the same, we study the series with the general term uk = a2kx2k and write (15.41) ALGEBRA OF SERlES 439 This gives us the limit Using the ratio test it is clear that this series converges for the interval (-1,l). However, at the end points the ratio test fails, where we now use the Raabe test. We first evaluate the ratio (15.43) = lim k [ ] = 5 > 1, (15.44) - l1 (2k + 2)(2lc + 1) lim k - - 1 = lim k k-cc [Uztt2 ] k+oo [ (2k)2-n2 6k+2+n2 3 k-co (21C)Z - n2 which indicates that the series is convergent at the end points as well. This means that for the polynomial solutions of the Chebyshev equation, restricting n to integer values is an additional assumption, which is not required by the finite solution condition at the end points. The same conclusion is valid for the series with the odd powers. 15.3.8 Alternating Series For a given series of the form Cr=’=, (-l),+’ a,, if a, is positive for all n, then the series is called an alternating series. In an alternating series for sufficiently large values of n, if a, is monotonic decreasing or constant and the limit lim a, = 0 n-cc (15.45) is true, then the series is convergent. This is also known as the Leibniz rule. Example 15.10. Leibnix rule: In the alternating series since $ > 0 and $ -+ 0 as n -+ 00, the series is convergent. 15.4 ALGEBRA OF SERIES Absolute convergence is very important in working with series. It is only for absolutely convergent series that ordinary algebraic manipulations (addition, subtraction, multiplication, etc.) can be done without problems: 1. An absolutely convergent series can be rearranged without affecting the sum. 440 INFINITE SERlES 2. Two absolutely convergent series can be multiplied with each other. The result is another absolutely convergent series, which converges to the multiplication of the individual series sums. All these operations look very natural; however, when applied to condi- tionally convergent series they may lead to erroneous results. Example 15.11. Conditionally convergent series: The following condi- tionally convergent series: 11 11 23 45 = 1 - (- - -) - (- - -) (15.47) = 1-0.167-0.05 , (15.48) obviously converges to some number less than one, actually to In 2 = 0.693. We now rearrange this sum as 11 1111 11 1 35 7 9 11 (If - +-)- ($+(- +- + - + E+ 15) - (2) 1 1 1 1 1 1 17 25 6 27 35 8 + (-++ +-) - (-) +(-++ *+ -) - (-) f , (15.49) and consider each term in parenthesis as the terms of a new series. Partial sums of this new series are ~1 = 1.5333, ~3 = 1.5218, ~2 = 1.0333, ~4 = 1.2718, Sg = 1.3853, $10 = 1.4078, S5 = 1.5143, Sg = 1.3476, . . . . (15.50) S7 = 1.5103, ~g = 1.5078, It is now seen that this alternating series added in this order converges to 3/2. What we have done is very simple. First we added positive terms until the partial sum was equal or just above 3/2 and then subtracted negative terms until the partial sum fell just below 3/2. In this process we have neither added nor subtracted anything from the series; we have simply added its terms in a different order. By a suitable arrangement of its terms a conditionally convergent series can be made to converge to any desired value or even to diverge. This result is also known as the Riemann theorem. 15.4.1 Rearrangement of Series Let us write the partial sum of a double series as (15.51) ALGEBRA OF SERIES 441 If the limit lim s,, = s 71'00 m-cc exists, then we can write (15.52) (15.53) and say that the double series CG=, aij is convergent and has the sum s. When a double sum (15.54) converges absolutely, that is, when Cz0C', laijl is convergent, then we can rearrange its terms without affecting the sum. Let us define new dummy variables q and p as i = q 2 0 andj =p - q 2 0. (15.55) Now the sum CEO ~~o aij becomes Writing both sums explicitly we get (15.56) (15.57) Another rearrangement can be obtained by the definitions 442 as 15.5 INFINITE SERIES r i=s>Oandj=r-2~>0, (ss $, (15.58) 0000 03 [a1 (15.60) - - a00 + a01 + a02 + +alo + ao3 + all + . . . USEFUL INEQUALITIES ABOUT SERIES Let + = 1; then we can state the following useful inequalities about series: H8lder’s Inequality: If an 2 0, b, 2 0, p > 1, then cx) c anbn 5 (2 u;) I”. (g b;) ‘Iq. (15.61) n= 1 n=l n=l Minkowski’s Inequality: If an 2.0, b, 2. 0 and p 2 1, then [ n= 2 1 (an + ha)’] I (2 n=l a;) + (5 n=l K) . (15.62) Schwarz-Cauchy Inequality: If a, 2 0, and bn 2 0, then (Fanbn)2 I (Fa:). n= 1 (Ebi). n=l (15.63) n= 1 Thus, if the series C,“==, a: and xT=l bi converges, then the series Cr=l anbn also converges. 15.6 SERIES OF FUNCTIONS We can also define series of functions with the general term un = un (z). In this case the partial sums Sn are also functions of z : S~(Z) = u~(z) +ua(z)+ +un(z). (15.64) If lim Sn(z) + S(Z) is true, then we can write n- n= 1 (15.65) [...]... in the interval -1 < x < 1 Note that for m = n (integer) the sum automatically terminates after a finite number of terms, where the quantity = m!/n!(m - n)!is called the binomial coefficient (z) Example 15.14 Relativistic kinetic energy: T h e binomial formula is probably one of the most widely used formulas in science and engineering An important application of the binomial formula was given by Einstein... effects like bending and twisting A practical use for the Casimir effect in our everyday lives is the pressure sensors of airbags in cars Casimir energy is bound to make significant changes in our concept of vacuum 15.11 INFINITE PRODUCTS Infinite products are closely related to infinite series Most of the known functions can be written as infinite products, which are also useful in calculating some of... INFINITE SERIES 455 we can write this integral as Using the following property of Bernoulli polynomials: (15 .133 ) Bi(2) = Bo(z)= 1, we can write Equation (15 .132 ) as I’ f(z)dz= s,’ (15 .134 ) f(z)Bo(z)dz = Integrating this by parts we obtain where we have used Bl(1) = use a a n d Bl(0) -4 .In t h e above integral we now = Blk) = 1 (15 .137 ) a n d integrate by parts again to obtain 1 1 f(z)dz= 5 [f(l)+ f(0)l-... be seen by taking Mi = /ail Sz in the M-test 15.8.2 Continuity In a power series, since every term, that is, u,(x) = unlcn, is a continuous function and since in the interval -S 5 2 5 S the series f(x) = C a n z n is uniformly convergent, f(x) is a continuous function Considering that in Fourier series even though the u ( )functions are continuous we expand dis,z continuous functions shaped like a... very important In 1999 Gerardus't Hooft and J G Martinus Veltman received the Nobel Prize for showing that Salam and Weinberg's theory of unified electromagnetic and weak interactions is renormalizable On the other hand, quantum gravity is nonrenormalizable because it contains infinitely many divergent pieces 15.10.1 Casimir Effect and Renormalization To demonstrate the regularization and renormalization... known series In this section we introduce some analytic techniques to evaluate the sums of infinite series We start with the Euler-Maclaurin sum formula, which has important applications in quantum field theory and Green’s function calculations Next we discuss how some infinite series can be summed by using the residue t h e orem Finally, we show that differintegrals can also be used to sum infinite series... I l ( z )and I2(x): (15.172) (15.173) In astrophysics we frequently work on gasses obeying the Maxwell-Boltzman distribution, where we encounter gamma functions defined as We now calculate I ( z , p ) for large values of x We first start by integrating the above integral by parts twice to get (15.175) and then (15.176) We keep on integrating by parts t o obtain the series SUMMATION OF INFINITE SERIES... numbers as B, 454 INFINITE SERIES 4 n 5 (15.124) 6 (15.125) 4, 1 7 In the interval [O, 1 and for s 2 1, the only zeroes of &+1(z) are 0, and 1 In the same interval 0 and 1 are the only zeroes of (Bzs(z)3 z S ) -l Bernoulli polynomials also satisfy the inequality IB2s(5)1 5 IBzsI, 0 I5 I 1 (15.126) 8 The Bernoulli periodic function, which is continuous and has the period one, is defined as P ( )= BSb... converging series with a rapidly converging one If we take t h e integral to the left-hand side, i t can be used for numerical evaluation of integrals 458 fNFlNITE SERIES AY (N+ 1/2) (-1 +I) I I -N-1 I I -N \ I I -2 CN -1 ( N + 1/2) (1 + i ) I 2 I I 1 N Y (N+ In) (-1- i ) I N+1 ) (N+ (1 -I) In) Fig 15.3 Contour for finding series sums using the residue theorem 15.9.3 Using Residue Theorem to Sum infinite... where the integral sign can be interchanged with the summation sign 3 If for all n in the interval [a, b] , ~ ( x ) -() and &x are continuous, and the series C" = ,= , is uniformly convergent, then we can differentiate the series term by term as gun(.) (15.74) (15.75) 15.7 TAYLOR SERIES Let us assume that a function has a continuous n t h derivative, f(n)(x), the in interval [a, b] Integrating this . energy: The binomial formula is prob- ably one of the most widely used formulas in science and engineering. An important application of the binomial formula was given by Einstein in his celebrated. (15.29) Let us find finite solutions of this equation in the interval z E [-1,1] by using the Frobenius method. We substitute the following series and its derivatives into the Chebyshev. us assume that a function has a continuous nth derivative, f(n)(x), in the interval [a, b]. Integrating this derivative we get Integrating again, - - f'"-2'(z) -