1 n. n > dt. t < n 1 + n=1


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1 Math 05 otes C. Pomerace The harmoic sum The harmoic sum is the sum of recirocals of the ositive itegers. We kow from calculus that it diverges, this is usually doe by the itegral test. There s a more elemetary roof that goes as follows: = = j=0 2 j+ > 2 j+ 2 = j+ 2, =2 j j=0 =2 j j=0 which diverges to ifiity. However, the itegral test is the better way because it ca rove fairly good uer ad lower bouds for the sum of the first N terms of the harmoic series. We have < t, 2 ad + > t,. Thus, addig these iequalities for u to N gets us N+ N t < N + = (The secod iequality treats the term = of the sum as just the umber, ad there s o estimate here, i fact, if N = the artial sum of the harmoic series is.) Evaluatig the itegrals gets us N log(n + ) < + log N. () = t. Losig a little iformatio here, this result could be writte as N = = log N + O(). This meas that there is some ositive costat c such that N log N c =
2 for all ositive itegers N. The Ootatio hides the costat c from view; it takes gettig used to! This result ca also be writte as = log x + O(), x. (2) x Here x is a real variable (we usually use ad N for iteger variables). I the iequality uder the summatio sig, it is imlicitly assumed that. A more recise way to write it is x = The reaso why the estimate holds is that the differece betwee log x ad log x is tiy, i fact it goes to 0 as x. So the additioal error made i writig log x istead of log x ca be absorbed ito the O() term i (2). This is all woderful, but ca we do better? That is, ca we be more recise about the error what ca be said about log N? N. Let a = + t. Itegratig ad usig roerties of log, we get a = ( log + ), ad exadig via the Taylor series for log( + x) which coverges for < x, we get a = ( 2 + ) = This is a alteratig series where the terms decrease i absolute value, so we see that 0 < a < 2. 2 A cosequece of this iequality is that the sum = coverges to a ositive costat. Call this costat γ. It is kow as Euler s costat or sometimes, the Euler Mascheroi costat. A very recet article about it ca be foud i the October 203 issue of the Bulleti of the America Mathematical Society, writte by Jeffrey Lagarias. To 0 decimal laces, γ = It is ot kow if γ is ratioal or irratioal. 2 a
3 Theorem. For x we have Proof. Let N = x. We have x x = N = log x + γ + O = N = N a + = a = N+ N+ ( ). x ( + a + t = log(n + ) + γ ) t a + log(n + ) N+ Let us estimate this remaiig sum usig the iequality 0 < a < above. This gives us 0 < a < 2. 2 N+ N+ a. 2 2 that we established This sum ca be majorized i two ways, oe more formulaic, the other more clever. forumlaic way is to use the iequality 2 < 2 2t, 2 so that N+ The clever way to estimate the sum is to use 2 < 2 N 2t = 2 2N. 2 < 2 2( ) = 2 N+ ( ) ad the use a telescoig series to get exactly the same estimate. I ay evet, this shows that ( ) ( ) a = O = O. N x To comlete the roof of the theorem we eed to show that ( ) log(n + ) = log x + O. x 3 The
4 But this is easy, usig the iequality log( + θ) θ, which holds for all θ > ad is easily roved usig calculus. (We did so i class.) Thus, if θ = (N + )/x, we have ( ) N + 0 < log(n + ) log x = log = log ( + θ) θ < x x. Above we itroduced Ootatio, see the book for a fuller descritio. We also itroduced Euler s costat, ad we set the toe for maiulatig sums ad aroximatig with the Taylor series. These tools will be commolace i the course. 2 The rime harmoic sum I 737, Euler roved that the sum of recirocals of the rimes (called the rime harmoic sum) diverges. I fact he showed that while the ordiary harmoic sum diverges like log, the rime harmoic sum diverges like loglog. See the recet article of Paul Pollack for more o what Euler kew or could have kow ( Euler ad the artial sums of the rime harmoic series at htt://www.math.uga.edu/ ollack/work.html). I articular, the sum of the recirocals of the rimes diverges. The book has a roof of the divergece of the rime harmoic series i Chater that we covered i class. Here s aother roof, also covered i class. Let P () deote the largest rime factor of whe 2, ad let P () =. So, for examle, sayig that P () 2 is just sayig that is of the form 2 j, ad sayig that P () 3 meas that = 2 j 3 k for some oegative itegers j, k. Note that P () 2 = j=0 2 j = 2. (The iformatio uder the first summatio sig here ecodes that is the dummy variable ad that it is ruig over all ositive itegers which satisfy the coditio.) It s more excitig to cosider P () 3 = j 0, k 0 2 j 3 k. This double sum over j ad k factors (usig the fudametal theorem of arithmetic) as j 0 2 j 3 = 2 3 k 2 = 3. k 0 More geerally, we have for ay iteger N 2, P () N = j 0 4 j =. (3)
5 Chagig the sum ito a roduct was a trick devised by Euler ad it deeds itimately o uique factorizatio ito rimes. The sum i (3) o ivolves all umbers N ad also lots of larger oes too, i fact ifiitely may. I ay evet, we have from (3) that N <. Usig that the sum is greater tha log N (see (2)), ad takig logs of this last iequality, gets us to log log N < ( ) log. (4) Agai usig the Taylor series, we have ( ) log = ( + log ) say. Let α deote the sum α = A = = = A, ( 2 + ) (The otatio idicates that we have a ifiite sum over all rimes.) We ote that the sum is coverget, sice 0 < A < ( 2 + ) = 3 2( ). (5) (Here, we relaced all of the umerical fractios i the terms i A with, makig the sum 2 larger, ad the saw a geometric series which ca be summed exactly.) So the series that defies α coverges with comariso to the series 2( ). We coclude from this that = ( ( )) log + ( ) log = ( ( )) log ( log >N ( ) log. = α >N A + ( )) + ( ) log 5
6 Usig the iequality (4) for the fial sum here, we have the that > log log N α A. >N The ifiite sum here is, from (5), less tha 0, but greater tha >N 2( ) > 2N. Sice the sum aears with a egative sig, we ca igore it i the iequality, gettig > log log N α. (6) So, the big questio ow is how accurate is this lower boud? Do we have a comaio uer boud? 3 A uer boud for the artial sums of the rime harmoic series We begi with the almost trivial estimate N log N N log = N log. The ier sum here is over the distict rime factors of. The ext ste is to iterchage the order of summatio, which will be a commo trick for us. This gets us to N log N log = N log > ( ) N log = ( ) N log log. N (7) Lemma (Erdős, Chebyshev). For each ositive iteger N, we have 4 N. Proof. It s true for N =, 2. If N is the least umber for which it fails, the N 3 ad N is odd (sice if N were eve it would be comosite, ad sice it is true for the revious odd = 6
7 umber, the roduct remais the same as with the revious odd umber). Write N = 2k +. We have the biomial coefficiet ( ) 2k+ k divisible by every rime i [k + 2, 2k + ], ad so ( ) 2k + = (( ) ( )) 2k + 2k + +, k 2 k k + k+2 2k+ sice the two biomial coefficiets here are equal. Their sum is smaller tha 2k+ j=0 Usig this i the above, we have ( ) 2k + = ( + ) 2k+ = 2 2k+. j k+2 2k+ < 2 22k+ = 2 2k = 4 k. Sice k + < 2k +, the lemma is true for k +, so we have = < 4 k+ 4 k = 4 2k+. 2k+ k+ k+2 2k+ Thus, N = 2k + is ot a couterexamle afterall, so the Lemma always holds. The logarithm of the iequality i Lemma is log N log 4. Usig this, movig some terms aroud i (7), ad dividig by N gets us log log N + log 4. (8) The issue ow is how to traverse from this estimate to a uer boud for the artial sum of the rime harmoic series. For this we itroduce the cocet of artial summatio. Proositio (Partial summatio). Suose that f(x) is a cotiuously differetiable real valued fuctio ad a for =, 2,... is a sequece of real umbers. The N a f() = f(n) N a N a f (t). t 7
8 Proof. The cotributio of a articular term o the left side is a f(). We comute what cotributes o the right side. After a little thought we see that this is f(n)a N a f (t) = f(n)a (a f(n) a f()) = a f(). The cotributios are the same, so we have roved the idetity. Partial summatio ca be used too whe we have a rime dummy variable, or more geerally for ay subsequece of the atural umbers. For examle, if oe wats to sum a f() ad use Proositio, we ca let a = 0 whe is ot rime, so the sum istatly gets trasformed to a f(), N sice all of the ew terms are 0. The sums o the right side of the idetity are also the same as if they were restricted to rimes, so we have a f() = f(n) a N 2 a f (t). Note the lower limit of itegratio is ow 2, sice we ca igore the = term, beig 0. We ow aly artial summatio to the sum i (8): Thus, usig (8), we have = log log = log N N (log N + log 4) + log N 2 = + log 4 log N log t + log 4 t log 2 t log + t N log 4 + log log N log log 2 log N + log 4 log 2 We have thus roved together with (2) that for x 2, x = log log x + O(). 2 log t log 2. t t = log log N + 3 log log 2. (9) 8
9 I metioed i class the theorem of Mertes from 874 that for x 2, ( ) = log log x + γ α + O. (0) log x x I class we worked out some exercises usig artial summatio. logarithm of the iequality i Lemma to show that π(x) = ( ) x = O. log x x I articular we used the We also showed that Mertes s theorem (0) imlies the same thig. Imrovig the error term slightly i (0) to o(/ log x) imlies, via artial summatio, the rime umber theorem, amely π(x) x log x as x. Sayig that the error i (0) is o(/ log x) meas that x (log log x + γ α) 0 as x. / log x Ad sayig that π(x) x/ log x as x meas that π(x) x/ log x as x. 9
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