RATIO MATHEMATICA ISSUE N. 30 (2016) pp. 59-66 ISSN (print): 1592-7415 ISSN (online): 2282-8214 The sum of the series of reciprocals of the quadratic polynomial with different negative integer roots Radovan Potůček 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] and [2] deal- ing with the sums of the series of reciprocals of some quadratic polynomials, deals with the series of reciprocals of the quadratic polynomials with differ- ent negative integer roots. We derive the formula for the sum of this series and verify it by some examples evaluated using the basic programming lan- guage of the CAS Maple 16. Keywords: sequence of partial sums, telescoping series, harmonic num- ber, computer algebra system Maple. 2010 AMS subject classifications: 40A05, 65B10 doi: 10.23755/rm.v30i1.12 1 Introduction and basic notions Let us recall the basic terms, concepts and notions. For any sequence {ak} of numbers the associated series is defined as the sum ∞∑ k=1 ak = a1 + a2 + a3 + · · · . 59 Radovan Potůček The sequence of partial sums {sn} associated to a series ∞∑ k=1 ak is defined for each n as the sum of the sequence {ak} from a1 to an, i.e. 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 telescoping series is any series where nearly every term cancels with a preceding or following term, so its partial sums eventually only have a fixed num- ber of terms after cancellation. Telescoping series are not very common in math- ematics but are interesting to study. The method of changing series whose terms are rational functions into telescoping series is that of transforming the rational functions by the method of partial fractions. For example, the series ∞∑ k=1 1 k2 + k has the general nth term an = 1 n(n + 1) = A n + B n + 1 . After removing the fractions we get the equation 1 = A(n + 1) + Bn. Solving it for A and B we obtain an = 1/n−1/(n + 1). After that we arrange the terms of the nth partial sum sn = a1 + a2 + · · · + an in a form where can be seen what is cancelling. Then we find the limit of the sequence of the partial sums sn in order to find the sum s of the infinite telescoping series as s = lim n→∞ sn. In our case we get sn = ( 1 1 − 1 2 ) + ( 1 2 − 1 3 ) +· · ·+ ( 1 n−1 − 1 n ) + ( 1 n − 1 n + 1 ) = 1− 1 n + 1 . So we have s = lim n→∞ ( 1− 1 n + 1 ) = 1. 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 . 60 The sum of the series of reciprocals of the quadratic polynomial The values of the sequence {Hn− lnn} decrease monotonically towards the limit γ . = 0.57721566, which is so-called the Euler-Mascheroni constant. Basic in- formation about harmonic numbers can be found e.g. in the web-sites [3] or [4], interesting information are included e.g. in the paper [5]. First ten values of the harmonic numbers are presented in this table: n 1 2 3 4 5 6 7 8 9 10 Hn 1 3 2 11 6 25 12 137 60 49 20 363 140 761 280 7129 2520 7381 2520 2 The sum of the series of reciprocals of the quadratic polynomial with different negative integer roots Now, we deal with the series formed by reciprocals of the normalized quadratic polynomial (k − a)(k − b), where a < b < 0 are integers. Let us consider the series ∞∑ k=1 1 (k −a)(k − b) , and determine its sum s(a,b). We express the nth term an of this series as the sum of two partial fractions an = 1 (n−a)(n− b) = A n−a + B n− b . From the equality of two linear polynomials 1 = A(n− b) + B(n−a) for n = a we get A = 1/(a−b) and for n = b we get B = 1/(b−a) = −1/(a−b). So we have an = 1 a− b ( 1 n−a − 1 n− b ) = 1 b−a ( 1 n− b − 1 n−a ) . (1) For the nth partial sum of the given series so we get sn = 1 b−a [( 1 1− b − 1 1−a ) + ( 1 2− b − 1 2−a ) + · · · · · ·+ ( 1 n−1− b − 1 n−1−a ) + ( 1 n− b − 1 n−a )] . The first terms that cancel each other will be obviously the terms for which for the suitable index ` it holds 1/(1−a) = 1/(`−b). Therefore the last term from the beginning of the expression of the nth partial sum sn, which will not cancel, will 61 Radovan Potůček be the term 1/(−a), so that the first terms from the beginning of the expression the sum sn, which will not cancel, will be the terms generating the sum 1 1− b + 1 2− b + · · ·+ 1 −a . Analogously, the last terms that cancel each other will be the terms for which for the suitable index m it holds 1/(n − b) = 1/(m − a). Therefore the first term from the ending of the expression of the nth partial sum sn, which will not cancel, will be the term 1/(n + 1 − b), so that the last terms from the ending in the expression of the sum sn, which will not cancel, will be the terms generating the sum − 1 n + 1− b − 1 n + 2− b −···− 1 n−a . After cancelling all the inside terms with the opposite signs we get the nth partial sum sn = 1 b−a ( 1 1− b + 1 2− b +· · ·+ 1 −a − 1 n + 1− b − 1 n + 2− b −···− 1 n−a ) . Because for integer c it holds lim n→∞ 1 n + c = 0, then the searched sum, where a < b < 0, is s(a,b) = lim n→∞ sn = 1 b−a ( 1 1− b + 1 2− b + · · ·+ 1 −a ) = = 1 b−a [ 1 1 + 1 2 + · · ·+ 1 −a − ( 1 1 + 1 2 + · · ·+ 1 −b )] , so we get Theorem 2.1. The series ∞∑ k=1 1 (k −a)(k − b) , where a < b < 0 are integers, has the sum s(a,b) = 1 b−a ( H−a −H−b ) , (2) where Hn is the nth harmonic number. Corolary 2.1. For the sum s(a,b) above it obviously hold: 1. s(a,b) = s(b,a), 2. s(a,a + 1) = H−a −H−a−1 = 1 −a , 62 The sum of the series of reciprocals of the quadratic polynomial 3. s(a,a+i) = 1 i ( H−a−H−a−i ) = 1 i ( 1 −a− i + 1 + 1 −a− i + 2 +· · ·+ 1 −a ) , i ∈ N. Remark 2.1. Let us note, that the formula (2) holds also in the case a < b = 0. Because H0 is defined as 0, it has the form s(a,0) = 1 0−a ( H−a −H0 ) = H−a −a . (3) Example 2.1. The series ∞∑ k=1 1( k − (−5) )( k − (−2) ) = ∞∑ k=1 1 (k + 2)(k + 5) , where a = −5, b = −2, has the nth partial sum sn = 1 3 ( 1 3 + 1 4 + 1 5 − 1 n + 3 − 1 n + 4 − 1 n + 5 ) . By the relation s(−5,−2) = lim n→∞ sn, since lim n→∞ 1 n + c = 0, or by Theorem 2.1 we get its sum s(−5,−2) = 1 3 ( 1 3 + 1 4 + 1 5 ) = 1 3 ( H5 −H2 ) = 1 3 ( 137 60 − 3 2 ) = 47 180 = 0.261. Example 2.2. The series ∞∑ k=1 1( k − (−5) ) k = ∞∑ k=1 1 k(k + 5) , where a = −5, b = 0, has the nth partial sum sn = 1 5 ( 1 1 + 1 2 + 1 3 + 1 4 + 1 5 − 1 n + 1 − 1 n + 2 − 1 n + 3 − 1 n + 4 − 1 n + 5 ) . By the relation s(−5,0) = lim n→∞ sn, since lim n→∞ 1 n + c = 0, or by Theorem 2.1, or by the Remark 2.1 we get its sum s(−5,0) = 1 5 ( 1 1 + 1 2 + 1 3 + 1 4 + 1 5 ) = H5 5 = 137/60 5 = 137 300 = 0.456 . 63 Radovan Potůček 3 Numerical verification We solve the problem to determine the values of the sum s(a,b) = ∞∑ k=1 1 (k −a)(k − b) for a = −1,−2, . . . ,−9 and b = a + 1,a + 2, . . . ,−8. We use on the one hand an approximative direct evaluation of the sum s(a,b,t) = t∑ k=1 1 (k −a)(k − b) , where t = 108, using the basic programming language of the computer algebra system Maple 16, and on the other hand the formula (2) for evaluation the sum s(a,b). We compare 45 = 9 + 8 + · · · + 1 pairs of these ways obtained sums s(a,b,108) and s(a,b) to verify the formula (2). We use following simple proce- dures hnum, rp2abneg and two for statements: hnum:=proc(h) local i,s; s:=0; if h=0 then s:=0 else for i from 1 to h do s:=s+1/i; end do; end if; end proc: rp2abneg:=proc(a,b,n) local k,sab,sumab; sumab:=0; sab:=(hnum(-a)-hnum(-b))/(b-a); print("n=",n,"s(",a,b,")=",evalf[20](sab)); for k from 1 to n do sumab:=sumab+1/((k-a)*(k-b)); end do; print("sum(",a,b,")=",evalf[20](sumab), "diff=",evalf[20](abs(sumab-sab))); end proc: for i from -1 by -1 to -9 do for j from i+1 by -1 to -8 do rp2abneg(i,j,100000000); end do; end do; 64 The sum of the series of reciprocals of the quadratic polynomial The approximative values of the sums s(a,b) rounded to 3 decimals obtained by these procedures are written into the following table: s(a, b) a =−1 a=−2 a=−3 a=−4 a=−5 a=−6 a=−7 a=−8 a=−9 b = 0 1.000 0.750 0.611 0.521 0.457 0.408 0.370 0.340 0.314 b=−1 × 0.500 0.417 0.361 0.321 0.290 0.266 0.245 0.229 b=−2 × × 0.333 0.292 0.261 0.238 0.219 0.203 0.190 b=−3 × × × 0.250 0.225 0.206 0.190 0.177 0.166 b=−4 × × × × 0.2000 0.183 0.170 0.159 0.149 b=−5 × × × × × 0.167 0.155 0.145 0.136 b=−6 × × × × × × 0.143 0.134 0.126 b=−7 × × × × × × × 0.125 0.118 b=−8 × × × × × × × × 0.111 Computation of 45 couples of the sums s(a,b,108) and s(a,b) took over 18 minu- tes. The absolute errors, i.e. the differences ∣∣s(a,b)−s(a,b,108)∣∣, have here place value about 10−8. 4 Conclusion We dealt with the sum of the series of reciprocals of the quadratic polynomials with different negative integer roots a and b, i.e. with the series ∞∑ k=1 1 (k −a)(k − b) , where a < b < 0 are integers. We derived that the sum s(a,b) of this series is given by the formula s(a,b) = 1 b−a ( H−a −H−b ) , where Hn is the nth harmonic number. We verified this result by computing 45 various sums by using the CAS Maple 16. We also stated that this formula holds also for a < b = 0, when it has the form s(a,0) = 1 0−a ( H−a −H0 ) = H−a −a . The series of reciprocals of the quadratic polynomials with different negative integer roots so belong to special types of infinite series, such as geometric and telescoping series, which sums are given analytically by means of a simple for- mula. 65 Radovan Potůček References [1] R. Potůček, The sums of the series of reciprocals of some quadratic poly- nomials. In: Proceedings of AFASES 2010, 12th International Conference ”Scientific Research and Education in the Air Force” (CD-ROM). Brasov, Romania, 2010, p. 1206-1209. ISBN 978-973-8415-76-8. [2] R. Potůček, The sum of the series of reciprocals of the quadratic polynomials with double non-positive integer root. In: Proceedings of the 15th Confer- ence on Applied Mathematics APLIMAT 2016. Faculty of Mechanical En- gineering, Slovak University of Technology in Bratislava, 2016, p. 919-925. ISBN 978-80-227-4531-4. [3] Wikipedia contributors: Harmonic number. Wikipedia, The Free Encyclopedia, [online], [cit. 2016-09-01]. Available from: https://en.wikipedia.org/wiki/Harmonic number. [4] E. W. Weisstein, Harmonic Number. From MathWorld – A Wol- fram Web Resource, [online], [cit. 2016-09-01]. Available from: http://mathworld.wolfram.com/HarmonicNumber.html [5] A. T. Benjamin, G. O. Preston, and J. J. Quinn, A Stirling En- counter with Harmonic Numbers. Mathematics Magazine 75 (2), 2002, p. 95 –103, [online], [cit. 2016-09-01]. Available from: https://www.math.hmc.edu/∼benjamin/papers/harmonic.pdf 66