k. s k=1 Part of the significance of the Riemann zeta-function stems from Theorem 9.2. If s > 1 then 1 p s

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1 9 Pimes in aithmetic ogession Definition 9 The Riemann zeta-function ζs) is the function which assigns to a eal numbe s > the convegent seies k s k Pat of the significance of the Riemann zeta-function stems fom Theoem 92 If s > then ζs) ) s Poof If we exand the RHS fo all of the imes u to x we get ) + s + ) s + 2s The oduct on the ight is a finite oduct, ove finitely many imes, of absolutely convegent geometic seies Thus we may eaange the tems of the sum in any convenient ode If we exand the oduct we then get As the seies ) s k: k k x + Σ x) + Σ 2 x) k k>x: k conveges, it follows that Σ x) conveges to ζs) and Σ 2 x) tends to zeo Eule imlicitly used the Riemann zeta-function to show that thee ae infinitely many imes by showing that the sum behind the LHS of 92) diveges at s one can make this agument igoous by taking the limit as s aoaches one fom above) Diichlet used the Riemann zeta-function to show that thee ae infinitely many imes in the aithmetic ogession an + b if and only if a and b ae coime We will do the secial case a 4 and b, we will show thee ae infinitely many imes of the fom 4n +

2 In fact he intoduced the notion of an L-function Ls) k also known as a Diichlet seies We will need a simle vaiant on the Jacobi symbol { 0 if 2 k /k) ) k )/2 k is odd We can use this to define define an L-function, /k) Ls) 3 + s 5 s 7 + s k Note that /) if and only if mod 4, and so it is not so suising that the Jacobi symbol tuns u Note that /k) is totally multilicative, so that /kl) /k) /l), fo all natual numbes k and l It follows that if k e e 2 2 e the ime factoisation of n then /n) / ) e / 2 ) e 2 / ) e We have a simle genealisation of 92): Lemma 93 If is a totally multilicative sequence of numbes meaning that the function k is totally multilicative) and the seies ak k s is conveges absolutely fo s > s 0 then ak k a s s fo s > s 0 ) Poof If we exand the RHS fo all of the imes u to x we get a ) + a ) s + a2 s + 2s The oduct on the ight is a finite oduct, ove finitely many imes, of absolutely convegent geometic seies Thus we may eaange the 2

3 tems of the sum in any convenient ode If we exand the oduct we then get a ) s As the seies k: k k x + Σ x) + Σ 2 x) k k>x: k conveges, it follows that Σ x) conveges to ζs) and Σ 2 x) tends to zeo Hee is a simle case of Diichlet s theoem: Theoem 94 Thee ae infinitely many imes q of the fom 4k + ; in fact thee ae so many that lim q s ) s + q Poof We need to undestand the behaviou of ζs) as s aoaches fom above We ae going to aly the summation fomula, 64), to λ n n, c n and ft) x s We have, fo s >, k x + s s x s k x x x s + s t dt ts+ t { t } t s+ dt x x s + s s x s ) s If we take the limit as x tends to infinity then we get ζs) s s s { t } dt ts+ In fact the integal conveges fo s > 0 It follows that lim s { t } dt ts+ +s )ζs) so that lim ζs) + 3 s

4 It is convenient to let q un ove the imes conguent to modulo 4 and fo to un ove the imes conguent to modulo 4 Since Ls) is eesented by a seies that is absolutely convegent fo s > we have Ls) /) ) s ) + ) q s s q Now conside the oduct ζs)ls) We have aleady shown that we can facto this into a oduct ζs)ls) ) ) 2 ), 2 s q s 2s q valid fo s > Conside what haens as s aoaches fom above Fist the RHS The fist tem aoaches 2) 2 Fo the last tem we have 0 < < 2s ) 2 ) 2 ) < ζ2) Thus the last tem is bounded We want the middle tem to go to infinity, so that the RHS goes to infinity Now conside the LHS We aleady showed that ζs) goes to infinity So if we can show that lim Ls) 0 s + then the LHS goes to infinity, which foces the RHS to go to infinity We will show that Ls) is continuous at s so that it suffices to show that L) 0 Recall that the seies u k s) is continuous, if u, u 2, ae continuous on a closed inteval I and they convege unifomly on the inteval I to the sum 4

5 In ou case u k s), ae continuous on the whole s-axis and so we just need to check unifomity on closed intevals [s, s 2 ], whee s > 0 We have to show that given ɛ > 0 thee is an m 0 such that n /k) < ɛ fo all m > m 0 and evey s I The idea is to aly Abel s atial summation fomula Note fist that n /k), egadless of m and n If we ut A m 0 and k A k /l) lm fo k m then n /k) n A k A k n A k k s k + ) s ) + A n n + ) s A m m s n k s k + ) s ) + n + ) s m s m s On the othe hand, m s < ɛ fo all m sufficiently lage Finally, note that L) /3 + /5 /7 + /9 + /3) + /5 /7) + /9 /) + > 2/3 5