RATIO MATHEMATICA 27 (2014) 69-79 ISSN:1592-7415 The sum of the series of reciprocals of the cubic polynomials with triple non-positive integer root Radovan Pot̊uček Department of Mathematics and Physics, Faculty of Military Technology, University of Defence, Brno, Czech Republic Radovan.Potucek@unob.cz Abstract This contribution, which is a follow-up to author’s paper [1] dealing with the sums of the series of reciprocals of some quadratic polynomi- als, deals with the series of reciprocals of the cubic polynomials with triple non-positive integer root. Three formulas for the sum of this kind of series expressed by means of harmonic numbers are derived and presented, together with one approximate formula, and verified by several examples evaluated using the basic programming language of the computer algebra system Maple 16. This contribution can be an inspiration for teachers who are teaching the topic Infinite series or as a subject matter for work with talented students. Key words: telescoping series, harmonic numbers, CAS Maple, Riemann zeta function. 2000 AMS: 40A05, 65B10. 1 Introduction Let us recall some basic terms. For any sequence {ak} of numbers the associated series is defined as the sum ∞∑ k=1 ak = a1 + a2 + a3 + · · · . The sequence of partial sums {sn} associated to a series ∞∑ k=1 ak is defined for each n 69 Radovan Pot̊uček as the sum sn = n∑ k=1 ak = a1 + a2 + · · · + an. The series ∞∑ k=1 ak converges to a limit s if and only if the sequence of partial sums {sn} converges to s, i.e. lim n→∞ sn = s. We say that the series ∞∑ k=1 ak has a sum s and write ∞∑ k=1 ak = s. The nth harmonic number is the sum of the reciprocals of the first n natural numbers: Hn = 1 + 1 2 + 1 3 + · · · + 1 n = n∑ k=1 1 k . The generalized harmonic numbers of order n in power r is the sum Hn,r = n∑ k=1 1 kr , (1) where Hn,1 = Hn are harmonic numbers. Every generalized harmonic number of order n in power m can be written as a function of generalized harmonic number of order n in power m− 1 using formula (see [2]): Hn,m = n−1∑ k=1 Hk,m−1 k(k + 1) + Hn,m−1 n , (2) whence Hn,2 = n−1∑ k=1 Hk k(k + 1) + Hn n , Hn,3 = n−1∑ k=1 Hk,2 k(k + 1) + Hn,2 n . Therefore Hn,3 = n−1∑ k=1 1 k(k + 1) (k−1∑ i=1 Hi i(i + 1) + Hk k ) + 1 n n−1∑ k=1 Hk k(k + 1) + Hn n2 , thus Hn,3 = n−1∑ k=1 1 k(k + 1) (k−1∑ i=1 Hi i(i + 1) + Hk k + Hk n ) + Hn n2 . (3) From formula (1), where r = 1, 2, 3 and n = 1, 2, . . . , 8, we get this table: n 1 2 3 4 5 6 7 8 Hn 1 3 2 11 6 25 12 137 60 49 20 363 140 761 280 Hn,2 1 5 4 49 36 205 144 5269 3600 5369 3600 266681 176400 1077749 705600 Hn,3 1 9 8 251 216 2035 1728 256103 216000 28567 24000 9822481 8232000 78708473 65856000 70 The sum of the series of reciprocals of the cubic polynomials 2 The sum of the series of reciprocals of the cubic polynomials with triple non-positive integer root We deal with the problem to determine the sum s(a,a,a) of the series ∞∑ k=1 1 (k −a)3 for non-positive integers a, i.e. to determine the sum s(0, 0, 0) of the series ∞∑ k=1 1 k3 = 1 13 + 1 23 + 1 33 + 1 43 + · · · , (4) the sum s(−1,−1,−1) of the series ∞∑ k=1 1 (k + 1)3 = 1 23 + 1 33 + 1 43 + · · · = s(0, 0, 0) −s1(0, 0, 0) = s(0, 0, 0) − 1 , the sum s(−2,−2,−2) of the series ∞∑ k=1 1 (k + 2)3 = 1 33 + 1 43 + · · · = s(0, 0, 0) −s2(0, 0, 0) = s(0, 0, 0) − 9 8 etc. Clearly, we get the formula ∞∑ k=1 1 (k −a)3 = s(0, 0, 0) −s−a(0, 0, 0) , (5) where s−a(0, 0, 0) is the (−a)th partial sum of the series (4). Several values of the nth partial sums sn(0, 0, 0), briefly denoted by sn, are: s100 . = 1.2020074, s1000 . = 1.2020564, s10000 . = 1.2020569, s100000 . = 1.2020569. Let us note that the series s(0, 0, 0) converges to the Apéry’s constant 1.202056903159 . . . , which represents the value ζ(3) of the Riemann zeta function ζ(3) = ∞∑ k=1 1 k3 = 1 13 + 1 23 + 1 33 + 1 43 + · · · . The partial sums sn(0, 0, 0) so present the generalized harmonic numbers Hn,3. According to formula (5) is s(a,a,a) = ζ(3) −H−a,3 , (6) then using formula (3) we get 71 Radovan Pot̊uček Theorem 2.1. The series ∞∑ k=1 1 (k −a)3 , where a is a negative integer, has the sum s(a,a,a) = ζ(3) − −a−1∑ k=1 1 k(k + 1) (k−1∑ i=1 Hi i(i + 1) + Hk k − Hk a ) − H−a a2 . (7) Now, we express formula (3) in another form. We have Hn,3 = 1 2 ( H1 1 + H1 n ) + 1 6 ( H1 2 + H2 2 + H2 n ) + + 1 12 ( H1 2 + H2 6 + H3 3 + H3 n ) + 1 20 ( H1 2 + H2 6 + H3 12 + H4 4 + H4 n ) + + 1 30 ( H1 2 + H2 6 + H3 12 + H4 20 + H5 5 + H5 n ) + · · · · · · + 1 (n− 3)(n− 2) ( H1 2 + H2 6 + · · · + Hn−4 (n− 4)(n− 3) + Hn−3 n− 3 + Hn−3 n ) + + 1 (n− 2)(n− 1) ( H1 2 + H2 6 + · · · + Hn−3 (n− 3)(n− 2) + Hn−2 n− 2 + Hn−2 n ) + + 1 (n− 1)n ( H1 2 + H2 6 + · · · + Hn−2 (n− 2)(n− 1) + Hn−1 n− 1 + Hn−1 n ) + Hn n2 , i.e. Hn,3 = H1 1 · 2 ( 1 1 + 1 2 · 3 + 1 3 · 4 + · · · + 1 (n− 2)(n− 1) + 1 (n− 1)n + 1 n ) + + H2 2 · 3 ( 1 2 + 1 3 · 4 + 1 4 · 5 + · · · + 1 (n− 2)(n− 1) + 1 (n− 1)n + 1 n ) + + H3 3 · 4 ( 1 3 + 1 4 · 5 + 1 5 · 6 + · · · + 1 (n− 2)(n− 1) + 1 (n− 1)n + 1 n ) + · · · · · · + Hn−3 (n− 3)(n− 2) ( 1 n− 3 + 1 (n− 2)(n− 1) + 1 (n− 1)n + 1 n ) + + Hn−2 (n− 2)(n− 1) ( 1 n− 2 + 1 (n− 1)n + 1 n ) + Hn−1 (n− 1)n ( 1 n− 1 + 1 n ) + Hn n2 . (8) Because 1 k(k + 1) = 1 k − 1 k + 1 , then the nth partial sum tn of the telescoping 72 The sum of the series of reciprocals of the cubic polynomials series ∞∑ k=2 1 k(k + 1) = 1 2 · 3 + 1 3 · 4 + · · ·+ 1 n(n + 1) + 1 (n + 1)(n + 2) + · · · is tn = ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) +· · ·+ ( 1 n − 1 n + 1 ) + ( 1 n + 1 − 1 n + 2 ) = 1 2 − 1 n + 2 , for the expressions in the first three parentheses of formula (8) we get 1 1 + 1 2 · 3 + 1 3 · 4 + · · · + 1 (n− 1)n + 1 n = 1 + tn−2 + 1 n = = 1 + ( 1 2 − 1 n ) + 1 n = 3 1 · 2 , 1 2 + 1 3 · 4 + 1 4 · 5 + · · · + 1 (n− 1)n + 1 n = 1 2 + tn−2 − t1 + 1 n = = 1 2 + 1 2 − ( 1 2 − 1 3 ) = 5 2 · 3 , 1 3 + 1 4 · 5 + 1 5 · 6 + · · · + 1 (n− 1)n + 1 n = 1 3 + tn−2 − t2 + 1 n = = 1 3 + 1 2 − ( 1 2 − 1 4 ) = 7 3 · 4 and analogously for the expressions in the last three parentheses of for- mula (8) we get 1 n−3 + 1 (n−2)(n−1) + 1 (n−1)n + 1 n = 1 n−3 + tn−2 − tn−4 + 1 n = = 1 n−3 + 1 2 − ( 1 2 − 1 n−2 ) = 2n−5 (n−3)(n−2) , 1 n−2 + 1 (n−1)n + 1 n = 1 n−2 + tn−2 − tn−3 + 1 n = = 1 n−2 + 1 2 − 1 2 + 1 n−1 = 2n−3 (n−2)(n−1) , 1 n−1 + 1 n = 2n−1 (n−1)n . Therefore Hn,3 = H1 1 · 2 · 3 1 · 2 + H2 2 · 3 · 5 2 · 3 + H3 3 · 4 · 7 3 · 4 + · · · · · · + Hn−2 (n−2)(n−1) · 2n−3 (n−2)(n−1) + Hn−1 (n−1)n · 2n−1 (n−1)n + Hn n2 , 73 Radovan Pot̊uček hence Hn,3 = 3H1 (1 · 2)2 + 5H2 (2 · 3)2 + · · · + (2n− 3)Hn−2 [(n− 2)(n− 1)]2 + (2n− 1)Hn−1 [(n− 1)n]2 + Hn n2 , thus Hn,3 = n−1∑ k=1 (2k + 1)Hk [k(k + 1)]2 + Hn n2 . (9) From formulas (6) and (9) we obtain Theorem 2.2. The series ∞∑ k=1 1 (k −a)3 , where a is a negative integer, has the sum s′(a,a,a) = ζ(3) − −a−1∑ k=1 (2k + 1)Hk [k(k + 1)]2 − H−a a2 . (10) Remark 2.1. In [4] it is derived that a good approximation for the partial sum sn(0, 0, 0) is the expression n∑ k=1 1 k3 ≈ ζ(3) − 1 4 ( 2 n2 − 2 n3 + 1 n4 ) . (11) It is stated that for small n, say, n = 5, the relative error in the above approximation is vanishingly small, i.e. about 0.03%, and that for larger n ∼ 1000, the error is swamped by machine precision. If we use formulas (11) and (5), where a is a negative integer, we get s(a,a,a) = s(0, 0, 0) −s−a(0, 0, 0) = = ζ(3) − −a∑ k=1 1 k3 ≈ ζ(3) − [ ζ(3) − 1 4 ( 2 (−a)2 − 2 (−a)3 + 1 (−a)4 )] , so we have an approximate formula s(a,a,a) ≈ 1 4 ( 2 a2 + 2 a3 + 1 a4 ) and an approximate sum s(a,a,a) = 2a2 + 2a + 1 4a4 . (12) Example 2.1. Evaluate the sum of the series ∞∑ k=1 1 (k + 5)3 74 The sum of the series of reciprocals of the cubic polynomials by means of formula: i) (7), ii) (10), iii) (5), iv) (12) and compare obtained results. Solution: i) The series has by Theorem 2.1, where a = −5, the sum s(−5,−5,−5) = ζ(3) − 4∑ k=1 1 k(k + 1) (k−1∑ i=1 Hi i(i + 1) + Hk k + Hk 5 ) − H5 25 . The last summand H5 25 = 137/60 25 = 137 1500 . Now, we evaluate the middle summand: 4∑ k=1 1 k(k + 1) (k−1∑ i=1 Hi i(i + 1) + Hk k + Hk 5 ) = 1 1 · 2 ( H1 1 + H1 5 ) + + 1 2 · 3 ( H1 1 · 2 + H2 2 + H2 5 ) + 1 3 · 4 ( H1 1 · 2 + H2 2 · 3 + H3 3 + H3 5 ) + + 1 4 · 5 ( H1 1 · 2 + H2 2 · 3 + H3 3 · 4 + H4 4 + H4 5 ) . If we denote this summand as S and use the values of the first five harmonic numbers from the table above, we get S = 1 2 ( 1 1 + 1 5 ) + 1 6 ( 1 2 + 3/2 2 + 3/2 5 ) + 1 12 ( 1 2 + 3/2 6 + 11/6 3 + 11/6 5 ) + + 1 20 ( 1 2 + 3/2 6 + 11/6 12 + 25/12 4 + 25/12 5 ) + = 1 2 · 6 5 + 1 6 · 31 20 + 1 12 · 311 180 + 1 20 · 265 144 = 1891 1728 . Altogether we have s(−5,−5,−5) = ζ(3) − 1891 1728 − 137 1500 = ζ(3) − 256103 216000 . = 0.016394866122 . ii) By Theorem 2.2 we get an easy and effective way how to obtain the required sum: s′(−5,−5,−5) = ζ(3) − 4∑ k=1 (2k + 1)Hk [k(k + 1)]2 − H5 25 = = ζ(3) − 3H1 (1 · 2)2 − 5H2 (2 · 3)2 − 7H3 (3 · 4)2 − 9H4 (4 · 5)2 − H5 25 . 75 Radovan Pot̊uček By means of the first five values of the harmonic numbers we have s′(−5,−5,−5) = ζ(3) − 3 4 · 1 − 5 36 · 3 2 − 7 144 · 11 6 − 9 400 · 25 12 − 1 25 · 137 60 = = ζ(3) − 256103 216000 . = 0.016394866122 . iii) The third and in this case much more easily way, how to determine the sum s(−5,−5,−5), is to use formula (5) and the value of s5(0, 0, 0) = H5,3 from the table above. So we immediately obtain the required result: s(−5,−5,−5) = s(0, 0, 0) −s5(0, 0, 0) = ζ(3) − 256103 216000 . = 0.016394866122 . iv) If we use formula (12), we get the approximate sum s(−5,−5,−5) = 2(−5)2 + 2(−5) + 1 4(−5)4 = 2 · 52 − 2 · 5 + 1 4 · 54 = 41 2500 = 0.0164 . Formulas (7), (10), and (5) give identical result 0.016394866122 , while for- mula (12) gives approximate result 0.0164 . The relative error of the fourth approximate result is 3.13 · 10−4 ∼ 0.03 %. 3 Numerical verification We solve the problem to determine the values of the sum s(a,a,a) of the series ∞∑ k=1 1 (k −a)3 for a = −1,−2, . . . ,−10,−99,−100,−500,−999,−1000. We use on the one hand an approximate evaluation of the sum s(a,a,a,t) = t∑ k=1 1 (k −a)3 , where t = 106, and formula (12) for approximate evaluation sum s(a,a,a), and on the other hand formulas (7) and (10) for evaluation the sum s(a,a,a) We compare 15 quadruplets of the sums s(a,a,a), s′(a,a,a), s(a,a,a, 106), and s(a,a,a) to verify formulas (7) and (10) and to determine the relative error of two approximate sums s(a,a,a, 106) and s(a,a,a). We use procedures hnum and rp3aaaneg written in the basic programming language of the CAS Maple 16 and one for statement: 76 The sum of the series of reciprocals of the cubic polynomials hnum:=proc(n) local m,h; h:=0; for m from 1 to n do h:=h+1/m; end do; end proc; rp3aaaneg:=proc(a,t) local i,k,A,A2,s,s1,s2,s3,saaa,s2aaa,sumaaa,sumaaaline,z3; A:=-a; A2:=A*A; s:=0; saaa:=0; s2aaa:=0; sumaaa:=0; z3:=1.20205690315959428540; for k from 1 to A-1 do s1:=0; s2:=0; s3:=0; if k-1=0 then s2:=0 else for i from 1 to k-1 do s2:=s2+hnum(i)/(i*(i+1)); end do; end if; s2:=s2+hnum(k)/k+hnum(k)/A; s1:=s1+s2/(k*(k+1)); s:=s+s1; s3:=s3+((2*k+1)*hnum(k))/(k*k*(k+1)*(k+1)); end do; saaa:=z3-s-hnum(A)/A2; s2aaa:=z3-s3-hnum(A)/A2; print("a=",a,":saaa=",evalf[20](saaa),s2aaa=",evalf[20](s2aaa)); for k from 1 to t do sumaaa:=sumaaa+1/((k-a)*(k-a)*(k-a)); end do; print("sumaaa(",t,")=",evalf[20](sumaaa)); sumaaaline:=(2*A2+2*a+1)/(4*A2*A2); print("sumaaaline=",evalf[20](sumaaaline)); print("rerrsumaaa=",evalf[20]((abs(sumaaa-saaa))/saaa)); print("rerrsumaaaline=",evalf[20]((abs(sumaaaline-saaa))/saaa)); end proc: A:=[-1,-2,-3,-4,-5,-6,-7,-8,-9,-10,-99,-100,-500,-999,-1000]; for a in A do rp3aaaneg(a,1000000); end do; The approximate values of the sums s(a,a,a) and s(a,a,a, 106), denoted briefly s and s(106), and the sum s(a,a,a), denoted s, obtained by the pro- cedures above and rounded to 9 decimals, are written into the following table (the sums s′(a,a,a) give identically values as the sums s(a,a,a)) : 77 Radovan Pot̊uček a −1 −2 −3 −4 −5 s 0.202056903 0.077056903 0.040019866 0.024394866 0.016394866 s(106) 0.202056903 0.077056903 0.040019866 0.024394866 0.016394866 s 0.250000000 0.078125000 0.040123457 0.024414063 0.016400000 a −6 −7 −8 −9 −10 s 0.011765236 0.008849784 0.006896659 0.005524917 0.004524917 s(106) 0.011765236 0.008849785 0.006896660 0.005524917 0.004524917 s 0.011766975 0.008850479 0.006896973 0.005525072 0.004525000 a −99 −100 −500 −999 −1000 s 0.000050502 0.000049502 0.000001996 0.000000501 0.000000500 s(106) 0.000050502 0.000049502 0.000001996 0.000000500 0.000000499 s 0.000050503 0.000049503 0.000001996 0.000000501 0.000000500 Computation of 15 quadruplets of the sums s(a,a,a), s′(a,a,a), s(a,a,a, 106) and s(a,a,a) took about 21 hours and 30 minutes. The relative errors of the approximate sums s(a,a,a, 106), i.e. the ratios∣∣[s(a,a,a, 106) −s(a,a,a)]/s(a,a,a)∣∣, range from 10−10 (for a = −1) to 10−5 (for a = −1000), and the relative errors of the approximate sums s(a,a,a), i.e. the ratios∣∣[s(a,a,a) −s(a,a,a)]/s(a,a,a)∣∣, range from 10−1 (for a = −1) to 10−5 (for a = −1000). 4 Conclusion We dealt with the sum of the series of reciprocals of the cubic polynomials with triple non-positive integer root a, i.e. with the series ∞∑ k=1 1 (k −a)3 . We stated that its sum clearly can be for great number of members t (we used t = 106) approximately computed by formula s(a,a,a,t) = t∑ k=1 1 (k −a)3 , we derived that the approximate value of its sum is for a negative a given by simple formula s(a,a,a) = 2a2 + 2a + 1 4a4 , 78 The sum of the series of reciprocals of the cubic polynomials and we derived that the precise value of the sum is for a negative a given by formula s(a,a,a) = ζ(3) −H−a,3 , i.e. by formula s(a,a,a) = ζ(3) − −a−1∑ k=1 1 k(k + 1) (k−1∑ i=1 Hi i(i + 1) + Hk k − Hk a ) − H−a a2 , and also by easier formula s′(a,a,a) = ζ(3) − −a−1∑ k=1 (2k + 1)Hk [k(k + 1)]2 − H−a a2 . We verified these results by computing 15 quadruplets of the four sums above for a = −1,−2, . . . ,−10, −99,−100,−500,−999,−1000 by using the CAS Maple 16 and compared their values. The series of reciprocals of the cubic polynomials with triple non-positive integer root so belong to special types of infinite series, such as geometric and telescoping series, which sums are given analytically by means of a formula which can be expressed in closed form. References [1] R. Pot̊uček, The sums of the series of reciprocals of some quadratic poly- nomials . In: Proceedings of AFASES 2010, 12th International Confer- ence ”Scientific Research and Education in the Air Force”. (CD-ROM). Brasov, Romania, 2010, 1206-1209. ISBN 978-973-8415-76-8. [2] Wikipedia contributors, Harmonic number . Wikipedia, The Free Encyclopedia, [online], [cit. 2015-06-25]. Available from: https://en.wikipedia.org/wiki/Harmonic number. [3] E. W. Weisstein, Harmonic Number . From MathWorld – A Wol- fram Web Resource, [online], [cit. 2015-06-15]. Available from: http://mathworld.wolfram.com/HarmonicNumber.html. [4] Mathematics Stack Exchange – A question and answer web- site for people studying math. [online], [cit. 2015-06-15]. Avail- able from: http://math.stackexchange.com/questions/361386/is-there- a-formula-for-sum-n-1k-frac1n3?rq=1. 79