DRAFT VOL 79, NO 1, FEBRUARY 26 1 The Euler-Maclaurin Forula and Sus of Powers Michael Z Spivey University of Puget Sound Tacoa, WA 98416 spivey@upsedu Matheaticians have long been intrigued by the su 1 + 2 + + n of the first n integers, where is a nonnegative integer The study of this su of powers led Jakob Bernoulli to the discovery of Bernoulli nubers and Bernoulli polynoials There are expressions for sus of powers in ters of Eulerian nubers and Stirling nubers [5, p 199] In addition, past articles in this MAGAZINE contain algoriths for producing a forula for the su involving powers of + 1 fro that involving powers of [1, 4] The algorith in Bloo is actually Bernoulli s ethod This note involves a curious property concerning sus of integer powers, naely, 1 + 2 + + 1 <, for 1 1 In other words, the su of the 1 ters fro 1 to 1 is always less than the single ter, regardless of how large is This inequality is not true for an arbitrary nuber of ters; 1 + 2 + + n 1 is not necessarily less than n for all n, but the inequality is true when n = Proving 1 is not too difficult In fact, one proof is a nice first-seester calculus proble using left-hand Rieann sus to underestiate the integral x dx Another establishes x + 1 x > x for x < via the binoial theore; replacing x successively with, 1, 2,, 1 and suing yields 1 There is a deeper question here, though Dividing 1 by produces the inequality 1 + + + < 1 2 Since this relation holds regardless of the value of, a natural question to ask is this: What is the liiting value of the expression on the left of 2 as approaches infinity? Our investigation of this value involves a useful tool in any atheatician s bag of tricks one that is, unfortunately, not often taught in undergraduate courses the Euler-Maclaurin forula for approxiating a finite su by an integral Along the way we also prove 1 using Euler-Maclaurin, thus illustrating the use of the Euler- Maclaurin forula with reainder Rota calls Euler-Maclaurin one of the ost rearkable forulas of atheatics [6, p 11] After all, it shows us how to trade a finite su for an integral It works uch like Taylor s forula: The equation involves an infinite series that ay be truncated at any point, leaving an error ter that can be bounded The forula uses the very nubers discovered by Bernoulli during his investigations into the power su, and the error ter uses Bernoulli s polynoials For exaple, the second-order forula with error ter is given in Concrete Matheatics [2,
2 MATHEMATICS MAGAZINE DRAFT p 469]: where n 1 f = = n + 1 3! fxdx + B 1 1! n fn f + B 2 f n f 2! B 2 x}f xdx, 3 B i is the ith Bernoulli nuber B 1 = 1/2, B 2 /6, B 2 x is the second Bernoulli polynoial: x 2 x + 1/6, and x} = x x Since x} is the fractional part of x, the function B 2 x} in 3 is ust the periodic extension of the parabola B 2 x = x 2 x + 1/6 fro [, 1] to the entire real nuber line In other words, B 2 x} agrees with B 2 x on [, 1] and is periodic with period 1 Proving 3 involves nothing ore coplicated than integration by parts A brief outline is as follows: Start with 1/2 1 y2 y + 1/6 g y dy Use integration by parts twice and solve for g Let gy = fy +, and then substitute x for y + to find an expression for f Su this expression as varies fro to n 1, noting that the ters involving f and f + 1 telescope, while those involving f + 1 are absorbed into the su This yields 3, since B 2 y} = B 2 x} The interested reader is invited to fill in the details The full Euler-Maclaurin forula with no reainder ter is given in Concrete Matheatics [2, p 471]: 1 = f = fxdx + B k f k 1 f k 1 4 Unfortunately, the infinite su on the right-hand side often diverges This forula can also be proved using integration by parts; Lapret, in fact, shows how to use parts to prove Euler-Maclaurin for arbitrary orders [3] On to the proof of 1: We can easily verify the inequality for sall values of In particular, for, we have < 1 1, and for = 2, we have < 4 = 2 2 For 3, we turn to Euler-Maclaurin Plugging fx = x and n = into 3 yields 1 =! x dx + 1 12 1 B 2 x} 1x 2 dx = +1 + 1 5 12 Need hypotheses on f Is this clear? I needed help when working this out B 2 x} 1x 2 dx 5
DRAFT VOL 79, NO 1, FEBRUARY 26 3 Now, let s deal with the error ter Copleting the square on the parabola B 2 x gives us B 2 x = x 1/2 2 1/12 This tells us that the iniu value of B 2 x on [, 1] is 1/12, occurring at x /2, and the axiu value on [, 1] is 1/6, occurring at the two endpoints x = and x Since B 2 x} is the periodic extension of B 2 x fro [, 1] to the real nuber line, the iniu and axiu values of B 2 x} over the real nubers are 1/12 and 1/6, respectively which, incidentally, occur infinitely often This tells us that 1/2B 2 x} 1/2 1/12 /24 Therefore, 1 2 B 2 x} 1x 2 dx 1 24 Plugging back into 5 produces 1 1x 2 dx = 24 1 24 +1 + 1 5 12 + 1 24 < 3 8 = 5 8 This establishes the inequality 1, naely 1 + 2 + + 1 <, for all positive integers, via the second-order Euler-Maclaurin forula with reainder We now ove on to our ain question deterining the liiting expression for 1 + + + Fro our proof of 1, we know that the liit ust be less than 5/8 To find the exact value we use the full Euler-Maclaurin forula 4 For fixed and fx = x, we have 1 1 = = + 1 + 1 x dx + 1 f k 1 Since f k 1 x is nonzero only for k + 1, this yields 1 = + 1 + 1 +1 f k 1 = + 1 + 1 +1 B k f k 1 f k 1 [ Bk k+1 1 k + 2 ] = +1 + 1 + [ Bk 1 k 1 k + 2 ]
4 MATHEMATICS MAGAZINE DRAFT There are exactly k 1 factors in the expression 1 k + 2 Thus the resulting polynoial is k 1 plus a polynoial of degree k 2 For our purposes, all that atters of the latter polynoial is its degree We can therefore use big-o notation to express 1 k + 2 as k 1 + O k 2 Here, O k 2 effectively eans that the expression added to k 1 is of order no larger than that of k 2 For a ore precise definition and a discussion of big-o notation, see Graha, Knuth, and Patashnik [2] Multiplying through by 1 k then yields the expression 1 + O 1/ Substituting back in, we have 1 Now we take the liit to get li 1 = = li + +1 + 1 + +1 + 1 + + li [ ] 1 1 + O O [ ] } 1 1 + O } +1 1 The crucial question for both the second and third ters is the convergence of k= / Fortunately, the infinite su is a special case of the exponential generating function for the Bernoulli nubers, k= x k = x e x 1, valid for x < 2π [5, p147] Therefore, +1 B k/ is bounded by a constant, yielding Since B, we have li li O 1 +1 1 = } k= =, which gives us the siple liiting expression [ ] 1 li + + + e 1 Thus, in the liit, the su 1 + 2 + + 1 will represent e 1 1 approxiately 582 of The interested reader ay enoy showing that the left-hand side of 2 actually increases to 1/e 1 In addition, the excellent text Concrete Matheatics contains nuerous further exaples of the use of the Euler-Maclaurin suation forula [2, pp 469 489] Acknowledgent The author would like to thank one of the referees for several helpful suggestions
DRAFT VOL 79, NO 1, FEBRUARY 26 5 REFERENCES 1 D M Bloo, An old algorith for the sus of integer powers this MAGAZINE, 66 1993, 34 35 2 R L Graha, D E Knuth, and O Patashnik, Concrete Matheatics, Addison-Wesley, Reading, MA, second edition, 1994 3 Vito Lapret, The Euler-Maclaurin and Taylor forulas: Twin, eleentary derivations, this MAGAZINE, 74 21 19 122 4 Robert W Owens, Sus of powers of integers, this MAGAZINE, 65 1992, 38 4 5 Kenneth H Rosen, ed, Handbook of Discrete and Cobinatorial Matheatics, CRC Press, Boca Raton, FL, 2 6 Gian-Carlo Rota, Cobinatorial snapshots, Math Intelligencer, 21 1999, 8 14