Elementary Analysis in Q p

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1 Elementary Analysis in Q Hannah Hutter, May Szedlák, Phili Wirth November 17, 2011 This reort follows very closely the book of Svetlana Katok 1. 1 Sequences and Series In this section we will see some roerties of sequences and series in Q. Recall from the first resentation that Q is a comlete metric sace. Hence every Cauchy sequence converges and therefore the set of the convergent sequences is the set of the Cauchy sequences. We will also see that we have some much better roerties in Q, than we are used to in the real case. The first examle is the characterization of the Cauchy sequences. Theorem 1. A sequence {a n } in Q is a Cauchy sequence (i.e. converges), if and only if lim n a n+1 a n = 0. Remark. This is obviously not true in R. For examle take a n = n 1 k=1 k 1 the n-th harmonic number. Then lim n a n+1 a n = lim n = 0 but n+1 as we know the a n is not Cauchy. Proof. Remember that if {a n } is a Cauchy sequence, then lim a m a n = 0. m,n The first direction follows immediately for m = n + 1. Note that this is true for Cauchy sequences in any metric sace. 1 Katok, Svetlana; -adic Analysis Comared with Real, American Mathematical Society,

2 For the converse assume lim n a n+1 a n = 0. This means that for any ɛ > 0 there exists N N s.t. for n > N a n+1 a n < ɛ. Then for any m > n > N, using the strong triangle inequality, we get a m a n = a m a m 1 + a m 1 a m 2 + a n max( a m a m 1,..., a n+1 a n ) < ɛ, which comletes the roof. Definition 1. Let the series a n be in Q. The sum converges if the sequence of its artial sums converges in Q, i.e. if lim N S N+1 S N = 0 where S N = N a n. The sum converges absolutely if a n converges in R. Examle. We have n 2 (n + 1)! = 2 in Q, for any. By induction we will show that for the N artial sum one has Then N n 2 (n + 1)! = 2 + (N 1) (N + 2)!. n 2 (n + 1)! 2 = lim 2 + (N 1) (N + 2)! 2 = 0 N since (N + 2)! 0 for N. The roof by induction is simle calculus: Case N = 1: 1 n 2 (n + 1)! = 1 2 = 2 = 2 + (1 1) (1 + 2)!, hence the induction base holds. For N N + 1: N+1 N n 2 (n + 1)! = n 2 (n + 1)! + (N + 1) 2 (N + 2)! =2 + (N 1) (N + 2)! + (N + 1) 2 (N + 2)! =2 + (N + 2)! (N 1 + N 2 + 2N + 1) =2 + N (N + 3)!. 2

3 As we are used to in R, convergence follows from absolute convergence by the triangle inequality. Proosition 1. If the series a i converges absolutely (in R), then a i converges in Q. Proof. Suose that a i converges. Then the sequence of its artial sums is Cauchy i.e. for any ɛ > 0 there exists N N s.t. for all m > n > N, we have m i=n+1 a i < ɛ. By the usual triangle inequality we get S m S n = m i=n+1 a i m i=n+1 a i < ɛ. Therefore {S n } is a Cauchy sequence and by definition a i converges in Q. We however can get a much better result in Q as seen in the next roosition. Proosition 2. A series a i in Q converges in Q if and only if lim i a i = 0 and in this case a i max a i. n Remark. The last roerty can be considered as the strongest wins roerty for converging series. Remark. We again know that this roosition is not true in R. By taking a i = 1 i, we can see that lim i a i = 0 but again the corresonding series does not converge. Proof. We know by definition that a i converges if and only if the sequence of artial sums S n = a i converges. Since a n = S n S n 1 the series converges if and only if a n goes to 0. For the second art we assume that a i converges. In the case where a i = 0 the roof is trivial. Now suose a i 0. Then by the strong triangle inequality for any artial sum we have n a i max a i. 1 i n Since a i 0 for n for large enough N we have max a i = max a n. 1 i N n 3

4 Since the artial sum n a i converges to a i 0 and the -adic norm can only take a discrete amount of values, for m large enough we have m =. The claim follows by taking the maximum of m a i and n. a i Definition 2. A series n=0 a n converges unconditionally if for any reordering of the terms a n a n the series n=0 a n also converges. Theorem 2. If n=0 a n converges, it converges unconditionally and the sum does not deend on the reordering. Remark. This is not true in R. We can even rove that for any convergent but not absolutely convergent series n=0 a n we have that for every x R there exists a reordering of the summands a n a n such that n=0 a n = x. Proof. Let ɛ > 0 and let N N s.t. for any n > N we have that a n < ɛ, a n < ɛ and N a n a n < ɛ. We know that we can find such an N since by Proosition 2 the terms a n tend to zero. Let S = N a n and S = N a n and define by S 1 and S 1, resectively, the sum of all terms in S for which a n ɛ and the sum of all terms of S for which a n ɛ. By construction it is clear that S 1 and S 1 contain the same terms and therefore S 1 = S 1. The sum S differs from S 1 only by the terms a n satisfying a n < ɛ and similarly S from S 1 by the terms a n satisfying a n < ɛ. Hence by alying the strong triangle inequality we get S S 1 < ɛ and S S 1 < ɛ and therefore S S < ɛ. Hence we have a n N a n = a n N a n + N a n N Now for ɛ 0 we see that the series a n converges and as desired. a n = a n a n < ɛ. 4

5 Remark. As in R, convergence does not imly absolute convergence. Consider the following series: 1; reeated times; 2 reeated 2 times;.... Since the terms converge to 0 the series converges but a n = =, and hence the series does not converge absolutely. Conclusion. Summing u the revious theorems and remarks we can conclude the following roerties. In Q one has the follwing imlications: absolute convergence convergence unconditional convergence. On the other hand in R one has: absolute convergence unconditional convergence convergence. Note that different from Q in R: convergence unconditional convergence. In both cases one has: convergence, unconditional convergence absolute convergence. The next roosition is retty subtle in the real case and stated here because it will be used later. Theorem 3. Let b ij Q, i, j = 1, 2,..., such that for any ɛ > 0 there exists N N such that max(i, j) N b ij < ɛ i.e. b ij tends to zero if either i or j goes to infinity. Then the two series ( ) b ij and j=1 ( ) b ij j=1 converge, and their sums are equal. Proof. We know from Proosition 2 that j=1 b ij converges for all i and b ij converges for all j. By the same roosition we know that for all i N and for all j N j=1 b ij max b ij < ɛ j b ij max i b ij < ɛ. 5

6 This imlies that both double series converge, again by Proosition 2. It is therefore left to show that both sums are equal. We know that = ( ) b ij j=1 N ( j=n+1 N ( N ) b ij b ij ) + { N ( max j=n+1 j=1 i=n+1 ) b ij, ( ) b ij j=1 i=n+1 ( ) } b ij < ɛ, where the last inequality follows from Proosition 2. Since this is true for any ɛ, the series must be equal. 2 -adic Power Series Definition 3. As in the real case a formal ower series is of the form f(x) = a n X n, n=0 where the coefficients a n are in Q and X is the indeterminate. The set of all ower series in X with coefficients in Q are denoted by Q [[X]] and the set of all olynomials by Q [X]. Let x Q and f Q [[ ]]. Then the corresonding ower series f(x) is n=0 a nx n. From Proosition 2 we know that f(x) converges if and only if a n x n 0. Therefore if for a 0 r R we have that a n r n 0, our series n=0 a nx n converges for any x Q s.t. x r. Definition 4. For a ower series f(x) = n=0 a nx n with a n Q the radius of convergence is defined as the extended real number 0 r f s.t. j=1 r f = su{r 0 : a n r n 0}. The extended real numbers are defined as R { }. The convergence radius can be comuted as in the Archimedian case. Proosition 3. For a ower series f(x) = n=0 a nx n the radius of convergence is 1 r f =. lim su n a n 1 n 6

7 Proof. The roof works as in the real case. Suose that x > r f (in the case where r f < ). We have lim su x a k 1 k = x lim su a k 1 k = x 1 r f > 1. Therefore for infinitely many values k we have that a k x k > 0 i.e. a k x k does not converge to 0 and hence the series diverges. Now suose that x < r f (in the case where r f > 0). Then we can choose r R such that x < r < r f. Then lim su r a k 1 k = r lim su a k 1 k = r r f < 1. Hence there exists an N N such that su r a k 1 k < 1. k N Therefore for k > N we have a k r k < 1 and ( ) k a k x k = a k r k x < x k 0 for k. r r k Since the terms of the series go to 0 it follows that the series converges. The natural question that arises is, whether the series converges on the boundary or not. Proosition 4. The domain of convergence of a -adic ower series f(x) Q [[X]] is a ball D = { x R} where R { k, k Z} {0} { }, and the series converges uniformly in D. Remark. We know that in the Archemedian case we can not make a statement about the convergence on the boundary. Take for examle the ower series of ln. We know that ln(1 + x) = ( 1)n+1 x n /n. One can find the the radius of convergence is 1, but for x = 1 the series diverges and for x = 1 the series converges. Proof. Suose the radius of convergence of f(x) is r f. As seen before the series converges on the oen ball D = { x < R}. We know that on the boundary i.e. where x = r f, f(x) converges if and only if a n x n 0. We can observe that this deends only on the norm of x and not the secific value of x, hence either for all oints with x = r f the series converges or it does for none. If the series converges for all oints with x = r f then we 7

8 have R = r f. If the series diverges for all oints with x = r f, then we have R = 1 r f. Finally for any x R we have a n x n a n R n 0 and therefore the uniform convergence on D follows. 3 -adic Logarithm and Exonential Definition 5. Similar as in the real setting, we define the -adic logarithm by the series ln (x) := ( 1) n+1 (x 1) n. n { } This series converges for all x B := x Z x 1 < 1 = 1 + Z. Note that the -adic logarithm corresonds to the formal ower series log (1 + X) = ( 1) n+1 X n n Q [[X]]. Theorem 4. The -adic logarithm satisfies the fundamental roerty ln (xy) = ln (x) + ln (y) for all x, y B such that xy is in B as well. Proof. Consider the following identity of formal ower series: log (1 + X) + log (1 + Y ) log (1 + X + Y + XY ) = 0. Clearly, this holds for the real logarithmic function, i.e. the left hand side of the above identity vanishes for all real numbers x, y ( 1, 1). We also know that all the series converge absolutely, which means that we can rearrange the terms such that the left side is of the form c i,j x i y j with c i,j = 0 for all i, j N. Now we choose any α, β Z. Clearly, α + β + αβ Z and by alying Theorem 3, the exression ln (1 + α) + ln (1 + β) ln ((1 + α) (1 + β)) can be written in the form c i,j α i β j where all c i,j = 0. This imlies the theorem. 8

9 Definition 6. Consider the formal ower series ex (X) = the corresonding ower series in Q is called the -adic exonential and is denoted by ex (x). Unfortunately, the radius of convergence of the -adic exonential is not as obvious as the one of the -adic logarithm. To find this, we need the following n=0 X n n! Lemma 1. For any ositive integer n N, let n = k a i i i=0 be its exansion in base. We then have where S n = a 0 + a a k. ord (n!) = n S n 1, Proof. If > n, we have S n = n and therefore n Sn = 0. Of course this is 1 true since n! and therefore ord (n!) = 0. Now let us assume n and let n = k i=0 a i i be the exansion of n in base. For any 1 i k there are exactly n i n i+1 numbers between 1 and n such that i is the is the greatest ower of dividing each. So we get ord (n!) = k ( n n ) i = i i+1 k n i. Here the last equation holds because for i 2, each term aears twice in the sum, once with coefficient i and once with coefficient (i + 1). Further, n = 0 by the definition of k. k+1 9

10 On the other hand we have ( n S n 1 = a0 + a a k k) (a 0 + a a k ) 1 = a ( 1 ( 1) + a 2 ( 2 1) a k k 1 ) 1 ( k k 1 ) = a i j = a i j j=0 1 j<i k = ( a 1 + a a k k 1) + ( a 2 + a a k k 2) a k n n n = + + = 2 k k n i as desired. Theorem 5. The -adic exonential ex converges in the disc } D := {x Q x < 1/( 1). Proof. The convergence radius r of a ower series a n X n is given by the formula 1 r =. lim su a n 1/n In our case, a n = 1/n!. Thus we get by Lemma 1 ord (r ) = lim inf 1 ( ) n n ord Sn (a n ) = lim inf n ( 1) = 1 1, where S n is the same as in Lemma 1 and the last equality holds since lim n S n n n Since r is a ower of, we have S n = 1 + lim n n = 1. r = ord(r) = 1/( 1). 10

11 Remark. If x = 1/( 1) then we have ( ) x n ord = n S n n! 1 + n 1 = S n 1. Now consider the subsequence n k = k. Then S nk = 1 and ord (x n k /nk!) = 1. Hence we have 1 lim x n k k n k! 0 for x = 1/( 1). This means that ex (x) diverges on D. Proosition 5. The -adic exonential satisfies the fundamental roerty for all x, y D. ex (x + y) = ex (x) ex (y) Proof. Similar as in Theorem 4, the result is true for formal ower series. Therefore we can conclude the result exactly the same way as in Theorem 4. Examle. Let = 2. Since 1 1 = 1/2 < 1, we know that 1 B = { } x Z x 1 < 1. So the 2-adic logarithm of 1 can be comuted: ( ) 2 1 ln 2 ( 1) = ln 2 (1 2) = On the other hand, we know from the fundamental roerty of the -adic logarithm that 0 = ln 2 (1) = ln 2 ( 1) + ln 2 ( 1) = 2 ln 2 ( 1). This means that the sum converges to 0 as n. 2 3 The next goal is to show that the -adic logarithm and the -adic exonential } are inverse to each other in some sense. Again, let D = {x Z x < r be the region of convergence of ex, where r = 1/( 1). Before we can rove the main result, we need the following estimates: Lemma 2. Let n 2 and let 0 < x < r. Then the following estimate holds: x n n x n n! < x < r. 11

12 Proof. From Lemma 1 we know that for any integer n 1, ord (n!) = n S n 1 n 1 1. But since ord (n) ord (n!), we clearly have n n! n 1 1 = r n 1. Therefore, by combining these two estimates, we get ( ) x n n x n n 1 x n! x < x < r. r Proosition 6. The mas ex : D 1 + D and ln : 1 + D D are inverse to each other. This means that for all x D, ln (ex (x)) = x and ex (ln (1 + x)) = 1 + x. Proof. By considering the corresonding formal ower series, we know that the relations in the Proosition hold. So the only thing we have to make sure is that all involved ower series converge. In other words, we have to show that ex (x) 1 + D and ln (1 + x) D for any x D. Then the convergence follows from the discussion above. By alying the estimate form the revious Lemma and the strongest wins roerty for series, we have ex (x) 1 = x + x n x n! < r. The last equality holds because all the terms in the sum are less than x by Lemma 2. Therefore, ex (x) 1 + D and ex : D 1 + D. The same argument works for the -adic logarithm. Let 1 + x 1 + D. Then ln (1 + x) = x + ( 1) n+1 x n x n < r. n=2 Again, the last equality follows from Lemma 2 and the strongest wins roerty for series. Thus we have ln (1 + x) D for all x D and therefore ln : 1 + D D. 12 n=2

13 } Now let x, y D and let z Z. Then x + y max { x, y zx x. Therefore, x + y and zx are in D as well, which means that D is an ideal in Z. In articular, D is an additive subgrou of Z. Further, x + y + xy D, which means that (1 + x) (1 + y) = 1 + x + y + xy 1 + D. This means that 1 + D is a multilicative subgrou of Z. Proosition 7. The -adic logarithm is an isomorhism of grous ex : D 1 + D. Proof. From the last Proosition, we already know that ex is bijective. Therefore it is enough to show that ex is a grou homomorhism. But this follows directly from the fundamental roerty: ex (x + y) = ex (x) ex (y). and Until now, the -adic logarithm and exonential behave quiet similar as the corresonding mas in the setting of real numbers. But there is one roerty which is totally different in the -adic setting. In fact, the -adic exonential is an isometry! Recall that an isometry is a ma that reserves the distance. To rove this, we need the following Proosition: Proosition 8. Let x D. Then the following holds: 1. ex (x) = 1, 2. ln (1 + x) = x, 3. 1 ex (x) = x. Proof. Let x D. By the estimate of Lemma 2, the term of maximal norm in the series x n ex (x) = 1 + x + n! is 1. By the strongest wins roerty for series we get that ex (x) = 1. Exactly the same way we see that in the series ex (x) 1 = x + n=2 n=2 x n n!, the term of maximal norm is x and therefore 1 ex (x) = x. 13

14 Similarly, the term of maximal norm in the series ln (1 + x) = x + is x and thus ln (1 + x) = x. Corollary 1. The mas ( 1) n+1 x n n=2 n are isometries. ex : D 1 + D and ln : 1 + D D Proof. Let x, y D. Then we have ex (x) ex (y) = ex (y) ex (x y) 1 = ex (x y) 1 = x y, where the second equation follows from art 1 and the last equation follows from art 3 of the revious roosition. This shows that ex is an isometry. To show the same for the -adic logarithm, we use that ex (ln (1 + x)) = 1 + x for x D. Then, since ln (1 + x) ln (1 + y) = ex (ln (1 + x)) ex (ln (1 + y)) = (1 + x) (1 + y) = x y, the -adic logarithm is an isometry. 4 Strassman s Theorem This last section is about the number of zeros of functions given by Z - converging ower series. First the Strassman s theorem is stated, which gives us an uer bound for the number of zeros for certain functions. Then some interesting corollaries will follow. Theorem 6 (Strassman s theorem). Let f : Z Q, x f(x), be given by a nonzero ower series f(x) = n 0 a nx n Q [[X]]. Suose that lim n a n = 0, so that f(x) converges for all x Z. Now let N N 0 be defined by: 1. a N = max n 0 a n, (since lim n a n = 0 a n attains its maximum for a finite set of indices). 14

15 2. a n < a N n > N, i.e., N is the largest index of a n with maximum norm. Now one can conclude that f has at most N zeros. Before roving Strassman s Theorem an examle: Examle. Consider the ower series f(x) = n 0 n!xn. We want (1) to determine the radius of convergence and (2) to estimate the number of zeros from above. (1) Set a n := n!. a n = 1 for all n {0,..., 1}, a n 1 for all n {,..., 2 1} and so on. So lim su n 0 a n = 1, hence the convergence radius is 1. Since a n 1 n n 0, f converges in Z. (2) The number we are looking for is N = 1. We are then able to conclude by alying Stassman that f has at most 1 zeros. Remark. In R this theorem is clearly not true. Consider to see this f(x) = sin(x) = n 0 ( 1) n (2n + 1)! X2n+1. The coefficients a n are given by { 0, if n even, a n = ( 1) n 1 2, n! else. From this, one can see that N = 1. But we know that sine has infinitely many zeros and not at most one! Proof. The roof is done by induction on N. N = 0: We know from the condition on N that a n > a 0 for all n > 0. Now assume for the urose of deriving a contradiction that there exists x Z such that 0 = f(x) = n 0 a n x n = a 0 + a 1 x + a 2 x We obtain that a 0 = n 1 a nx n, which can be rewritten as the following since we can use the strongest wins -roerty for sums: Finally we get (since x 1): a 0 = n 1 a n x n max n 1 a nx n. a 0 max n 1 a nx n max n 1 a n < a 0, 15

16 which is a contradiction. Our assumtion was roved wrong so there exists no x Z such that f(x) = 0. This shows the base of induction. N 1: Assume the statement of the theorem holds for m < N. We show that it also holds for N. To do this we factor out one zero and show that we are allowed to use the induction hyothesis on our new ower series. Assume that f has at least one zero in Z denoted by α. Otherwise the roof is trivial (0 < N). Suose N is given as in the assumtion of the theorem. Choose x Z arbitrary. We get f(x) = f(x) f(α) = n 0 a n (x n α n ) = (x α) n 0 where the last equality can be justified by multilying out n 1 (x α) x j α n 1 j = x n α n. j=0 n 1 a n x j α n 1 j, Since x and α are elements of Z, we have x, α 1. Hence we obtain that a n x j α n 1 j n a n 0. We can use Theorem 3(interchanging sums) and achieve that where f(x) = (x α) n 1 a n x j α n 1 j = (x α) n 0 j=0 j 0 = (x α)g 1 (x), g 1 (x) = j 0 b j x j, b j = n=j+1 a n α n 1 j = j=0 n=j+1 a n x j α n 1 j a j+1+k α k. Now we show that (1) g 1 converges in Z and (2) the largest index of b j with maximal norm is N 1. j (1) g 1 converges in Z if b j 0 (theorem from before). This is indeed the case (again with the strongest wins -roerty): b j = k=0 k=0 a j+1+k α k max k 0 a j+1+kα k max a j j+1+k 0. k 0 (2) We have b j max k 0 a j+1+k a N j N 0, 16

17 To obtain b N 1 = a N one has to work a bit more. We have: b N 1 = k 0 a N+k α k = lim n n a N+k α k, where the last equality holds because of continuity of the norms. By the strongest-wins -roerty from the first talk, i.e., if a < b then a + b = b and since a N+k α k a N+k < a N k 1, we have that b N 1 = lim n k=0 n a N+k α k = a N. k=0 j N : b j max k 0 a j+1+k = max n N+1 a n < a N. So b N 1 has maximal norm and N 1 is the largest index with maximal norm. We are now allowed to use the induction hyothesis on g 1, hence g 1 has at most N 1 zeros. We conclude that f has at most N = N zeros. So now we are able to derive some corollaries: Corollary 2. Let f again be given by a nonzero ower series f(x) = n 0 a n X n, that converges in Z. Say α 1,..., α m are all the zeros of f(x) in Z. Then there exists a ower series g(x) converging in Z with no zeros in Z and it holds that: f(x) = (X α 1 ) (X α m )g(x). Proof. From the roof of Strassman s Theorem we know that we can construct g 1 (X) a ower series in Z [[X]] with at most m 1 zeros and f(x) = (X α 1 )g 1 (X). Alying the same rocedure to g 1 we get g 2 with at most m 2 zeros and f(x) = (X α 1 )(X α 2 )g 2 (X). Reeating this rocess we can find g m with no zeros and Setting g := g m finishes the roof. f(x) = (X α 1 ) (X α m )g m (X). 17

18 Corollary 3. Let f(x) = n 0 a nx n be a nonzero ower series converging in m Z for some m Z. Then f(x) has finitely many zeros in m Z. The number of zeros is bounded from above by N, where N satifies mn a N = max n 0 mn a n and mn a n < mn a N n > N. Proof. Define g(x) := f( m X) = n 0 a n mn X n. As f converges in m Z, g converges in Z. We show that the number of zeros of g in Z is the same as the number of zeros of f in m Z : If x Z is a zero of g, then m x is a zero of f in m Z. Conversely if y m Z is a zero of f, then m y is a zero of g in Z. If we aly Strassman to g and the result follows immediately. Corollary 4. Let f(x) = n 0 a nx n and g(x) = n 0 b nx n two -adic nonzero ower series converging in m Z. Assume there exist infinitely many numbers α m Z such that f(α) = g(α), then a n = b n for all n N 0. Proof. Define h(x) = f(x) g(x) converging in m Z. If h were nonzero, it would have a finite number of zeros in m Z by Corollary 3. But it has infinitely many zeros in m Z as all the α s are zeros of h. So h has to reresent the zero ower series. Therefore we get a n = b n for all n N 0. Remark. Corollaries 2 and 4 have a real resectively comlex counterart. Corollary 5. Let f(x) = n 0 a nx n be nonzero and convergent in m Z. If f(x) is eriodic, i.e., there exists τ m Z such that f(x + τ) = f(x) for all x m Z, then f(x) is constant. Remark. This is not at all what we exect and know in real analysis. For examle we know that sine and cosine are eriodic ower series converging everywhere on R but are not at all constant. The difference comes from the fact that for ower series in R if τ is a eriod, then nτ for n Z do not belong to a bounded interval (but in -adic analysis they do). Proof. Set h(x) = f(x) f(0). h has zeros at τn for all n Z because f is eriodic. Since τ m Z and m Z is an ideal, we have that nτ m Z. So h has infinitely many zeros in m Z. By Corollary 4 this leads us to h(x) = 0 and thus f(x) = f(0) = const. Corollary 6. Let f(x) = n 0 a nx n be a nonzero ower series converging for all x Q. Then f(x) has at most a countable set of zeros. If the set of n zeros {z n } n 1 is not finite, then it forms a sequence with z n. 18

19 Proof. We know from the second and third talk that if x Q then there exists m Z such that x m Z. So we get that {x Q f(x) = 0} m Z{x m Z f(x) = 0}. From Corollary 3 we know that in m Z, f has only finitely many zeros. So the set of all zeros in Q is a countable union of finite sets, thus countable. Let the set of zeros not be finite and assume that su n 1 z n C <. Then there exists m Z such that {z n } n 1 m Z and by Corollary 3 n {z n } n 1 is finite, which is the desired contradiction. So z n. 19