POWER SUMS, BERNOULLI NUMBERS, AND RIEMANN S. 1. Power sums


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1 POWER SUMS, BERNOULLI NUMBERS, AND RIEMANN S ζfunction.. Power sus We begin wih a definiion of power sus, S (n. This quaniy is defined for posiive inegers > 0 and n > as he su of h powers of he firs n inegers: n S (n = i = (n. i= Our arge is o find an explicie closed forula for S (n in ers of and n. We already know soe special cases of he forula in quesion. In paricular, i is easy o see ha S ( =, and i is no difficul o verify he following forulas using aheaical inducion S (n = S (n = n(n, n(n (n, 6 S 3 (n = n (n. 4 Alhough i is no difficul o prove he above forulas if hey are saed, i does no see easy o guess he exac shape of he. One can, however, ake a general guess, which we forulae and prove as a proposiion. Proposiion. S (n ay be represened as a polynoial in n of degree +. The consan er of his polynoial is zero, and he leading er of his polynoial is n + /( + Proof. Recall he sandard binoial forula: for a posiive ineger d, one has ( ( d d (l + k d = l d + l d k + l d k k d. We will need a special case of his forula. Naely, we pu d = +, l =, and subsruc he er k + fro boh sides o obain ( ( ( ( + k + k + = + k + k k. Le us now plug in he values k = 0,,,..., n ino he laer ideniy, and wrie down he n forulas obained in his way:
2 + 0 + = + + = + ( ( ( = + ( ( ( + n + (n + = + ( + (n + ( + (n ( + Le us now su hese n ideniies ogeher, observing he cancellaions in he lefhand side and he appearence of power sus in he righ: ( ( ( n + = n + S (n + S (n S (n. Taking ino he accoun ha ( + = + we obain finally ( ( + + (+S (n = n + S (n S (n... ( + S (n Exercise. Finish he proof of Proposiion using he laer ideniy and a aheaical inducion arguen. Exercise. Use he proof of Proposiion in order o find he forula for S 5 (n. (n.. Bernoulli nubers Alhuogh Proposiion provides valuable qualiaive inforaion abou he forulas for S (n, and he proof of his proposiion acually allows o find hese forulas, we will now presen anoher way o wrie hese forulas down. In order o do so we define he sequence B 0, B, B... of Bernoulli nubers as he coefficiens of he Maclaurin series for he funcion /(e : e = B!. One can reforulae he above forula by saying ha he funcion /(e is he generaing funcion for he sequence of raional nubers B /!. A sraighforward bu edious calculaion of he Maclaurin series allows us o find he firs few Bernoulli nubers: B 0 =, B =, B = 6, B 3 = 0, B 4 = 30, B 5 = 0. B 6 = 4,... We now ake Proposiion ore precise by esablishing a connecion beween power sus and Bernoulli nubers.
3 3 Theore. (Bernoulli For an inegers n > and > 0 ( + ( + S (n = B j n + j. j j=0 Proof. We begin wih a sandard Maclaurin series for he exponenial funcion: e x x =!. We pu x = k in he above forula, an obain e k = k!. We now wrie he laer forula down for k = 0,,,..., n : = e =! e =! e (n = (n!. We add hese n ideniies, and observe he appearance of he power sus in he righhand side: ( + e + e e (n = + S (n! Transfor he lefhand side using he geoeric series forula + e + e e (n = en e = en e, noice ha boh facors have Maclaurin series e n = n k k k! e = j B j j!, and uliply hese wo series e n k= e = =k+j k j 0 j=0 n k B j k!j! = + S (n!.
4 4 Equae like powers of in he above ideniy and obain S (n = n k B j! k!j!. =k+j k j 0 In order o finish he proof we uliply boh sides of his ideniy by ( +!, and use he obvious fac ha ( ( +! + ( + j!j! =. j Exercise 3. Prove ha B n+ = 0 for n. Hin. The funcion /(e + / is even. Exercise 4. Define he sequence of raional nubers b 0, b, b,... as follows. Pu b 0 = 0, and for ( + ( + b = b k. k k=0 Prove ha b = B. Hin. In he definiion of Bernoulli nubers, uliply boh sides by e, and wrie he Maclourin series in for his funcion. Equae like coefficiens of like powers of, and show ha Bernoulli nubers saisfy he above ideniy. Explain, why his fac iplies b = B. Show ha i suffices o prove ha Bernoulli nubers B saisfy he above ideniy. This exercise presens an alernaive definiion for Bernoulli nubers. 3. Rieann s ζfuncion Rieann s ζfuncion is a funcion on s defined by he series ζ(s = n s. n= Honesly, his series is considered for a coplex variable s. We do no do ha here. Insead, for a posiive ineger, we consider he values ζ( = n = n= This series obviously converges (why?, and ay be aken as an analogue of he power sus. The sus, however are now infinie, and he exponens are negaive, hus he analogy is no close. This akes even ore ipressive he following classical resul which connecs hese infinie sus o Bernoulli nubers Theore. (Euler ζ( = ( +(π (! B.
5 5 Proof. The proof requires he following ideniy fro real analysis cox = x x n π x. n= We ake his ideniy for graned. We will now ake use of he infinie geoeric series forula in order o ransfor he suands in he righhand side: We hus obain xcox = x n π x = n= l= x n π (x/nπ = x n π ( x l x l = nπ π l l= n= l=0 ( x l. nπ n l = x l ζ(l. πl Now le us ransfor x co x. We need soe basic facs fro coplex analysis, naely cosx = eix + e ix and, sin x = eix e ix i which we also ake for graned. Thus xcox = ix eix + e ix e ix e ix = ixeix + e ix = ixeix + e ix = ix+ ix e ix = ix+ B (ix! We hus have wo power series (in x represenaions for he funcion xcox, and can equae he. Afer ha we equae like powers of x, and finish he proof of he heore. l=
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