In the last few sections we have seen several functions which have been defined via series or integrals. We now want to develop tools that will allow us to show that these functions are analytic. Recall that in general, it is not enough to know that the sum
f(x) = lim fn(x) n→∞
converges everywhere and that each fn(x) is differentiable to deduce that f is differentiable, sin(n2x) or that if f is differentiable, that f ′(x) = lim f ′ (x). For example, if f (x) = n→∞ k n n then f (x) f(x) := 0 for all x R, but f ′ (x) f ′(x)=0. n → ∈ n 6→ We’ll see that things behave a little more nicely for analytic functions because of the convergence properties of power series. Recall that the above is asking whether
f (z + δz) f (z) f (z + δz) f (z) lim lim n − n = lim lim n − n , δz→0 n→∞ δz n→∞ δz→0 δz
so what we are trying to do is swap the order of two limits. Similarly, asking whether
lim fn(z) dz = lim fn(z) dz n→∞ n→∞ Z Z is a question of whether two limits can be swapped, as the integral is also a kind of limit. In order to do this sort of limit swapping you need some extra hypotheses!
Definition: Let f1, f2, f3, ... be a sequence of complex-valued functions defined on a com- mon domain S C. Then f f uniformly on S (or f converges to f uniformly on S) if ⊆ n → n for all ǫ > 0 there is an integer N such that for every n N, f (z) f(z) < ǫ whenever ≥ | n − | z S, that is, such that sup fn(z) f(z) ǫ. ∈ z∈S | − | ≤ This is to be constrasted with the weaker notion of pointwise convergence: f f n → pointwise on S means that for all z S, lim fn(z)= f(z), or in full, ∈ n→∞
ǫ> 0 z S N n N f (z) f(z) < ǫ. ∀ ∀ ∈ ∃ ∀ ≥ | n − |
In this definition N = N(ǫ, z), that is, N depends on both the choice of ǫ and z, whilst in the definition of uniform convergence N = N(ǫ), that is, N depends on ǫ alone.
18 Recall: Uniform convergence is a metric space convergence, using the metric
d∞(f,g)= f g ∞ = sup f(z) g(z) . k − k z∈S | − | Of course one must limit this to functions where this quantity is defined, such as the set of all bounded functions on S. (The definition of uniform convergence doesn’t require that the functions are bounded, but one has to take great care if this is not the case!) Pointwise convergence is not (in general) a metric space convergence, which is partly why it is less well-behaved. The following result shows that uniform limits and contour integration commute.
Lemma 11. For n = 1, 2, 3,... let fn be a continuous function on a region Ω and suppose that f f uniformly on Ω. Then f is continuous on Ω and, for any contour γ lying in Ω, n →
lim fn(z) dz = f(z) dz. n→∞ Zγ Zγ Proof. The continuity of f follows from the standard ǫ/3 argument that you saw in earlier courses, using the fact that for any z,w Ω and any n, ∈ f(z) f(w) f(z) f (z) + f (z) f (w) + f (w) f(w) . | − |≤| − n | | n − n | | n − | Furthermore standard integral bounds tell us that
fn(z) dz f(z) dz fn(z) f(z) dz fn f ∞ length γ. γ − γ ≤ γ − | |≤k − k · Z Z Z
and so f (z) dz f(z) dz n → Zγ Zγ as f f 0. k n − k∞ →
Example: The fact that a contour has finite length is important here! Let fn(z) = 2 2 e−(z−n) /n for n =1, 2, 3,... and for 1 < Im z < 1. Then f f(z) = 0 uniformly on the √n − n → strip. However ∞ π ∞ f (z) dz = = f(z) dz. n 2 6 Z−∞ r Z−∞ The reason the above proof doesn’t work in this case is that length γ = . ∞ As we saw above, in the real setting, even the taking of uniform limits and differentiation do not commute. However, in the complex setting, the previous lemma plus Morera’s theorem gives the following result.
19 Theorem 12. For n =1, 2, 3,... suppose f is analytic in a disk z z < R and suppose n | − 0| f f uniformly in each closed disk z z R δ, where δ > 0. Then f is analytic in n → | − 0| ≤ − z z < R, and, moreover, f ′ f ′ uniformly in each disk z z R δ. | − 0| n → | − 0| ≤ −
Proof. Given Morera’s theorem it is enough to show that γ f(z) dz = 0 whenever γ is, say, a triangular contour in z z0 < R. For such a contourR choose δ > 0 small enough so | − | that γ is contained in z z R δ. Then | − 0| ≤ −
f(z) dz = lim fn(z) dz =0 n→∞ Zγ Zγ where the first equality follows by the previous lemma and the second equality follows from Cauchy’s theorem. We have proved that f is analytic in z z < R. | − 0| Next assume z z R δ and let C denote the circle z z = R δ/2. By Cauchy’s | − 0| ≤ − | − 0| − formula for deriviatives we have
1 dζ f ′(z) f ′ (z)= f(ζ) f (ζ) − n 2πi − n (ζ z)2 ZC − Now given ǫ> 0 we can choose N so that, for all ζ C, ∈ ǫδ2 f(ζ) f (ζ) < whenever n N. | − n | 4R ≥
Therefore, for n N, ≥ 1 ǫδ2 2πR δ f ′(z) f ′ (z) since ζ z for ζ C. | − n | ≤ 2π 4R (δ/2)2 | − | ≥ 2 ∈
In other words f ′(z) f ′ (z) < ǫ for n N. Since z is an arbitrary point satisfying | − n | ≥ z z R δ (and N does not depend on z) we have proved the result. | − 0| ≤ −
The restriction of the region to a disk can be relaxed in the above result. We will frequently ∞ be interested in the (uniform) convergence of series of functions n=1 fn on an more general set. Roughly speaking if the series converges uniformly on someP compact set, then so does ∞ ′ its series of deriviatives (assuming the functions are analytic) n=1 fn, and moreover
∞ d ∞ P f ′ = f . n dz n n=1 n=1 ! X X ∞ Recall: fn is defined to be the limit lim sN where sN = f1 + f2 + + fN is the Nth n=1 N→∞ ··· partial sumP of the series of functions.
20 Theorem 13. Suppose that f ∞ is a sequence of analytic functions defined on a region { n}n=1 Ω C. Suppose that ∞ f converges uniformly on each compact subset of Ω with limit ⊆ n=1 n ∞ ′ ′ function f. Then f isP analytic on Ω and moreover n=1 fn converges uniformly to f on each compact subset of Ω. P
Proof. This result follows very directly from the previous theorem. Suppose that K is a compact subset of Ω. For each z K, there exists δ(z) > 0 such ∈ that the disk D(z, 2δ(z)) = w : z w < 2δ(z) Ω. Clearly D(z, δ(z)) : z K forms { | − | } ⊂ { ∈ } an open cover of K, so by compactness, one can choose a finite subcover:
K D(z , δ(z )) D(z , δ(z )). ⊆ 1 1 ∪···∪ k k
Let Dj denote the closure of D(zj, δ(zj)), j =1,...,k. The series converges uniformly to f on each of these closed disks, so by the previous theorem, f is analytic on each of the open disks, and hence on K. Furthermore s′ f ′ uniformly on each of the closed disks D N → j Since there are only finitely many of these disks covering K, one may therefore deduce that s′ f ′ uniformly on K. N → The most commonly involved test to guarantee uniform convergence is the Weierstrass M- test.
Theorem 14. (Weierstrass M-test) Suppose that S is a nonempty subset of C and that f ∞ is a sequence of functions defined on S. Suppose that for each n, there exists a { n}n=1 constant M such that f M . If ∞ M < then the series ∞ f converges n k nk∞ ≤ n n=1 n ∞ n=1 n uniformly on S. P P
Recall: This theorem just a version of the Banach space theorem that if X is a Banach space and x is a sequence in X with x < , then x convergences in the Banach space. { n} k nk ∞ n Exercise: Check you could proveP this! P Recall: A series of complex numbers c converges absolutely if c converges. As in n | n| the real numbers, absolute convergenceP implies convergence. A seriesP of complex functions f converges absolutely on S if f (z) converges absolutely for every z S n | n | ∈ ∞ Exercise:P (i) Show that under theP hypotheses of the Weierstrass M-test, the series n=1 fn converges absolutely as well as uniformly. P
21 (ii) Give an example of a series of analytic functions on a set S which converges uniformly, but not absolutely. (iii) Give an example of a series of analytic functions on a set S which converges absolutely, but not uniformly. sin n2z Exercise: Let f (z)= , n =1, 2, 3,... . Then f f 0 uniformly on the real line. n n n → ≡ Furthermore, each f entire, as is the limit function. So why doesn’t f ′ f ′ on the real n n → line?
∞ n Theorem 15. If n=1 anz has radius of convergence ρ > 0 (meaning that the series con- ∞ n verges, indeed convergesP absolutely for z < ρ) then anz converges uniformly in any | | n=1 disk z R where R < ρ. P | | ≤
Proof. Assume ξ satisfies R< ξ < ρ then a ξn converges so that lim a ξn = 0. Hence | | n n→∞ n there is a constant M such that P
M a n =1, 2,.... | n| ≤ ξ n | | But then, for all z satisfying z R, we have | | ≤ M a zn Rn, neither M nor ξ depends on z. | n | ≤ ξ n | | | | But R/ ξ < 1 so that the geometric series ∞ (R/ ξ )n converges. Hence, by the Weier- | | n=1 | | n strass M-test anz converges uniformly forP z < R. | | P
Remark: Note that we need not have uniform convergence on z < ρ. (For example, | | 1/(1 z) = zn, z < 1 pointwise but not uniformly on z < 1). Nevertheless we can − | | | | conclude, byP Theorem 13, that
d ∞ ∞ a zn = na zn−1 z < ρ. dz n n | | n=1 ! n=1 X X Example: Show that ∞ 1 1 z + n − n n=−∞X, n6=0 is uniformly and absolutely convergent on each compact subset K of the punctured plane C n : n Z . \{ ∈ }
22 Solution. Since K is compact, K is bounded and so there is a constant M such that z M for all z K. Therefore | | ≤ ∈ 1 1 z M 1 = for all z K. z + n − n n(z + n) ≤ n2 1+ z/n ∈
| | Now for n > 2M, 1+ z/n > 1 /2 whilst | | | | inf 1+ z/n = δ > 0, z K, n 2M | | ∈ | | ≤ since K is compact, K is closed which means C K is open. Therefore there exists a constant \ A such that 1 1 A for all n Z and z K. z + n − n ≤ n2 ∈ ∈
P A But 2 is convergent. Hence the result follows by the Weierstrass M-test. n6=0 n Since each function a (z)= 1 1 is analytic on C n : n Z it follows that n z+n − n \{ ∈ } ∞ 1 1 z + n − n n=−∞X,n6=0 is analytic on C n : n Z and moreover, by Theorem 3, that \{ ∈ } d ∞ 1 1 ∞ 1 = − 2 . dz z + n − n ! (z + n) n=−∞X,n6=0 n=−∞X,n6=0 We have shown that we can differentiate this sum by differentiating each term.
23 5 Infinite products
5.1 Interpolation
A common and classical problem in calculus is to find a function that takes specified values at certain specified points. For example, it is easy to find a polynomial p for which you have specified N values
p(zj)= cj, j =1,...,N.
Indeed you can choose the Lagrange interpolating polynomial,
N (z zi) p(z)= c i6=j − . j (z z ) j=1 Qi6=j j i X − Q Written another way, this expresses p(z) as cjℓj(z) where ℓj is a polynomial for which
ℓj(zi)= δij. P If you wish to find an analytic function that satisfies infinitely many conditions, things are much more complicated. We have already seen, for example, that it is impossible to find an entire function f such that
0, if j is even f(1/j)= (1/j, if j is odd. (It is of course easy to construct a continuous function with these constraints!) A natural question to ask is whether one can ever construct anything like the Lagrange interpolating polynomial with infinitely many terms. This would require us to write some form of infinite product of terms (z z ). This obviously requires some care as such an − j infinite product can easily be zero or unbounded!
One of the amazing truths about analytic functions is that essentially all entire functions can be written as infinite products. This was discovered but not proved, as was so much else, by Euler in about 1750. We shall begin this section by studying one example:
sin πz ∞ z2 = 1 . πz − n2 n=1 Y This example introduces most of the key ideas. We shall then look more carefully at the theory of infinite products of numbers, such as
2 ∞ 1 = 1 π − 4n2 n=1 Y 24 and infinite products of functions. Later we shall apply the ideas to two further examples, the gamma function and the zeta function.
5.2 A nontrivial example
The basis of our example is the following partial fraction series representation of cot πz.
1 ∞ 2z π cot πz = + (for z =0, 1, 2,...). z z2 n2 6 ± ± n=1 X − To prove this we consider the contour CN which is the boundary of the rectangle bounded by the lines y = N, y = N, x = N + 1 and x = N + 1 . − 2 − 2
y
x = N + 1 x = N + 1 − 2 y = N 2 1 0 1 N N +1 −bb b b b x
y = N −
We first show that on C there is a bound for cot πz which is independent of N, that N | | is, there is a B such that, for all N =1, 2, 3,... cot πz B whenever z C . Note that | | ≤ ∈ N
cos πz = cos πx cosh πy i sin πx sinh πy − sin πz = sin πx cosh πy + i cos πx sinh πy.
Therefore, on x = (N +1/2), ± cos πz sinh πy sinh πy | | = | | = | | 1 sin πz cosh πy 2 ≤ | | | | 1 + sinh πy p
25 whilst on y = N, ± cos πz cos2 πx cosh2 πN + sin2 πx sinh2 πN | | = sin πz 2 2 2 2 | | psin πx cosh πN + cos πx sinh πN pcos2 πx + sinh2 πN = psin2 πx + sinh2 πN p1 + sinh2 πN 1 1+ ≤ s sinh2 πN ≤ sinh πN 1 1+ < 2. ≤ πN So in fact we can take B = 2. Suppose now that z / Z and make sure that N > z . We will now evaluate ∈ | | 1 π cot πw I = dw. N,z 2πi w2 z2 ZCN − Note that from the above bounds, and using the fact that if w C , then w2 N 2 and ∈ N | | ≥ hence w2 z2 N 2 z 2, | − | ≥ −| | 1 π cot πw 1 1 I | | dw dw = 2(4N + 1), | N,z| ≤ 2π w2 z2 | | ≤ N 2 z 2 | | N 2 z 2 · ZCN | − | ZCN −| | −| | which tends to 0 as N . →∞ On the other hand, for any N, the integral is just the sum of the residues at the singu- larities inside C . The singularities occur where sin πw = 0 and where w2 z2 = 0, that is, N − at integers w = N, (N 1),..., (N 1), N and w = z. π−cot−πw − − ± Let f(w) = . Then each integer is a simple zero of sin πw and thus is a simple w2 z2 pole of f and the Laurent− expansion is of the form c f(w)= −1 + c + c (w n)+ c (w n)2 + w n 0 1 − 2 − ··· − Calculating the residue is easy at such a point: π(w n) cos πw 1 Res(f, n) = lim (w n)f(w) = lim − = . w→n − w→n (w2 z2) sin πw n2 z2 − − A similar argument shows that the residues at z and z are − π cot πz π cot( πz) and − 2z 2z − Thus the Residue Theorem gives
N π cot πz π cot( πz) N 1 I = Res(f, z) + Res(f, z)+ Res(f, n)= + − + N,z − 2z 2z n2 z2 n=−N n=−N X − X − 26 Thus, with z fixed, and taking a limit as N , →∞ π cot πz π cot( πz) ∞ 1 lim IN,z =0= + − + N→∞ 2z 2z n2 z2 n=−∞ − X − or, using symmetry and pulling out the n = 0 term,
π cot πz 1 ∞ 2 0= + . z − z2 n2 z2 n=1 X − Therefore, multiplying through by z,
1 ∞ 2z π cot πz = − z z2 n2 n=1 X − as claimed. Clearly each term on the right hand side has an antiderivative related to log(z2 n2) and − the left hand side has an antiderivative related to log(sin πz) log(πz). If no-one is looking − you might write that
sin πz ∞ ∞ log(sin πz) log(πz) = log = log(z2 n2) = log (z2 n2) − πz − − n=1 n=1 X Y or sin πz ∞ = (z2 n2) πz − n=1 Y . . . before you realize that you have no idea what any of this means! Our task then is to turn this into something that does make sense to make this rigorous. sin πz Let us begin by considering ℓsin(z) = Log on the punctured disk, 0 < z < 1. πz | | Lemma 16. ℓsin(z) is analytic on the punctured disk, 0 < z < 1, where, as usual, Log is | | the principal branch of log.
Proof. Log f(z) is analytic in any region where the analytic function f(z) is neither zero nor takes a negative real value, since this is the cut for the function Log. First note that sin πz/z can only vanish where sin πz = 0, and none of the zeros is inside the punctured disk. Next suppose that the function takes a negative real value, k, or sin πz = kπz − − where k > 0. Equating real parts gives sin πx cosh πy = kπx or simply sin πx = k′πx − −
27 where k′ > 0.
y
f(x) = sin πx
1 2 x g(x)= k′πx, k′ > 0 −
From the graphs this can only occur if x = 0 or x > 1. So we can assume that x = 0 | | and equate the imaginary parts to obtain sinh πy = kπy. Since k > 0 the only solution is − y = 0 and therefore the function ℓsin is analytic on the punctured disk.
In fact, the function sin πz/πz clearly has a removable singularity at z = 0 and so we can make ℓsin analytic at z = 0, provided we give sin πz/πz the value 1 there. Consequently ℓsin is analytic on the disk z < 1. Moreover, for 0 < z < 1 we have | | | | d d sin πz d 1 ℓsin z = Log = (Log sin πz Log z Log π)= π cot πz . dz dz πz dz − − − z
From this we deduce that the deriviative of ℓsin at z = 0 must equal lim (π cot πz 1 )=0. z→0 − z Note also that 2 −2z d z n2 2z Log 1 2 = z2 = 2 2 . dz − n 1 2 z n − n − Thus, for z < 1, | | d ∞ 2z ℓsin z = dz z2 n2 n=1 X − ∞ d z2 = Log 1 dz − n2 n=1 X d ∞ z2 = Log 1 . dz − n2 n=1 X
28 Here we could interchange d/dz and since the series of Logs converges uniformly (on any compact set not containing an integer.)P One easy way to prove this is to observe that Log(1 w) 2 w when w 1/2. Therefore | − | ≤ | | | | ≤ z2 2 z 2 2 Log 1 | | for z < 1 and n 2. − n2 ≤ n2 ≤ n2 | | ≥
∞ 2 But 2 < and so we have the result by the Weierstrass M-test (we can clearly forget n=2 n ∞ a finiteP number of terms in the series if we want to). Therefore, for z < 1, | | sin πz ∞ z2 Log = Log 1 + C πz − n2 n=1 X for some constant C. Substituting z = 0 shows that C = 0. That is,
sin πz ∞ z2 N z2 Log = Log 1 = lim Log 1 . πz − n2 N→∞ − n2 n=1 n=1 X X At this point we would like to take the sum inside the logariPi*xthm, but we need to remember that Log a + Log b is not necessarily Log ab. Rather
Log a + Log b = ln a + i Arg a + ln b + i Arg b = ln ab + i(Arg a + Arg b) | | | |
If Arg a + Arg b ( π, π) then everything does work OK. ∈ − 2 z2 −1 |z| Exercise: (a) Show that Arg 1 2 sin 2 . − n ≤ n (b) Use the fact that t 1.1 sin t for t [0, 0.25] to show that for z < 1 and n 2, ≤ ∈ | | ≥ z 2 1.1 sin−1 | | < . n2 n2 (c) Deduce that for any N,
N z2 π N 1 π π2 Arg 1 < +1.1 +1.1 1 < 2.3 < π. − n2 2 n2 ≤ 2 6 − n=1 n=2 X X Pi*x In light of this then,
sin πz ∞ z2 N z2 Log = Log 1 = lim Log 1 πz − n2 N→∞ − n2 n=1 n=1 X X N z2 = lim Log 1 N→∞ − n2 n=1 ! Y
29 Now recalling that exp(Log w)= w for all w = 0, and, of course, that exp is continuous, 6 sin πz sin πz = exp Log πz πz N z2 = exp lim Log 1 N→∞ − n2 n=1 !! Y N z2 = lim exp Log 1 N→∞ − n2 n=1 !! Y N z2 = lim 1 . N→∞ − n2 n=1 Y We shall write this as sin πz ∞ z2 = 1 πz − n2 n=1 Y for z < 1. We will soon see that this equation holds for all z. | | If we substitute z =1/2 we obtain
1 ∞ 1 ∞ (2n 1)(2n + 1) = 1 = − π − 4n2 (2n)2 2 n=1 n=1 Y Y so that π ∞ (2n)2 π 22 42 62 = or = 2 (2n 1)(2n + 1) 2 1 3 3 5 5 7 ··· n=1 Y − · · · which is traditionally and rather unsatisfactorily written as Wallis’ product
π 22 42 62 = · · ··· . 2 12 32 52 72 · · · ··· Our immediate aim is to provide a general theoretical framework for what we have done, to regularize the example and to allow us to treat other examples.
5.3 Products of constants
∞ Definition: If pn n=1 is a sequence of non-zero complex numbers we say that the infinite ∞ { } product p converges to P if the sequence of partial products P = p p p converges n N 1 2 ··· N n=1 to a non-zeroY limit P . If the infinite products converge to zero or infinity then the product is said to diverge. Infinite products which do not converge are said to diverge.
Remark: Intuitively, for the infinite product to converge we’d expect to need that Log pn converges to 0, which means that p 1. As we’ll see shortly, this turns out to be correct, n →
30 so it is common to consider infinite products of the form ∞ (1 + α ) where α = 1 for n=1 n n 6 − all but a finite number of n. Clearly we then must have αQn 0 for the product to exist. → ∞ Definition: More generally we agree to say that an infinite product pn exists if n=1 Y 1. at most a finite number of factors are zero; and
2. the product of the non-vanishing terms exists in the above sense.
Hence a (convergent) infinite product has the value 0 if and only if one or more of its factors is 0. Example: (a) Consider ∞ 1 1 2 3 1 =0 − n · 2 · 3 · 4 ··· n=1 Y so that the infinite product exists if and only if lim 1 2 n−1 exists and is non-zero. n→∞ 2 3 ··· n 1 ∞ 1 But the value is = limn→∞ = 0. Therefore the infinite product (1 ) does not exist. n 1 − n It diverges to 0. (Ignoring the initial 0, this corresponds to the factQ that
N n log = log n +1 n +1 − n=1 X diverges to .) −∞ Example: (b) Consider
∞ ( 1)n 1 1 1 1 4 3 6 1+ − = 1 1+ 1 = . n +1 − 2 3 − 4 ··· 2 · 3 · 4 · 5 ··· n=1 Y N+2 Here PN =1/2 if N is odd and = 2(N+1) if N is even. Clearly then limN→∞ PN =1/2 so that the infinite product exists and equals 1/2. Note that if 0 were prefixed as the first factor in example (b) then the product would still exist and would equal 0. We say that it converges to 0. ∞ ∞ Example: (c) It is tempting to think that if P = n=1 pn exists then Log P = n=1 Log pn. This need not be the case however. Q P Let p = exp( iπ/2n), so that the partial products are given by n − N P = p = exp( iπ(1 1/2N )). N n − − n=1 Y ∞ In this case P P = 1, so we have Log p = πi while Log P = πi. N → − n − n=1 X 31 Lemma 17. If P = ∞ p exists then p 1. n=1 n n → Q Proof. If P exists, but is zero, then we need only look at those terms past the last pn that is zero, so without loss of generality, let’s assume that P = 0. 6 With the notation as above then,
Pn P lim pn = lim = =1. n→∞ n→∞ Pn−1 P
∞ ∞ Theorem 18. P = n=1 pn exists and is nonzero if and only if n=1 Log pn converges. Q P Proof. As we saw in the concrete example we did earlier, the main issue here is that we might not have Log pn = Log pn. Recall however that for any N
Q N P N Arg p = Arg p +2k π, some k Z, n n N N ∈ n=1 ! n=1 Y X and so N N
Log pn = Log pn +2kN πi. n=1 n=1 Y X Thus
N N N N
PN = pn = exp Log pn = exp Log pn +2kN πi = exp Log pn . n=1 n=1 ! n=1 ! n=1 ! Y Y X X One direction is now easy! If ∞ Log p converges then P exp ( ∞ Log p ) as n=1 n N → n=1 n N (since exp is continuous)P and so PN is convergent with a non-zeroP limit. →∞ 1 For the converse , assume that P = lim PN exists and is nonzero. Let us first assume that
P ( , 0). Then Log is continuous at P and hence Log P = Log lim PN = lim Log PN 6∈ −∞ N N exists. As above, for all N,
N
cN := Log PN = Log pn +2kN πi n=1 X for some integer k . Since the sequence c converges it is Cauchy, and in particular N { N }
c c = Log p +2πi(k k ) 0, as N . N − N−1 N N − N−1 → →∞ 1Thanks to Joel for googling the fix to this proof!
32 But we know that p 1, and hence that Log p 0. This implies that 2πi(k k ) 0 N → N → N − N−1 → too, which can obviously only happen if the sequence k is eventually constant. Thus, { N } there exists an integer k such that for all sufficiently large N,
N Log p = Log P 2kπi. n N − n=1 X Since the right-hand side converges (to Log P 2kπi), the left-hand side converges too. − The remaining situation is if P ( , 0). If every p were real and positive, then P ∈ −∞ n would be too, so there must exist at least one term, say pm which is not real and positive. ′ In this case P = pn converges to a nonzero point which is not on the negative real axis. n6=m Y ∞
The previous case would then imply that Log pn converges, and hence that Log pn n6=m n=1 converges too. X X ∞
Remark: It is worth noting that we actually showed that if pn is non-zero then the sum n=1 ∞ Y n=1 Log pn can only differ from Log P by an integer multiple of 2πi. P The following result provides us with a very simple test.
Theorem 19. ∞ Log p converges absolutely if and only if ∞ (1 p ) converges ab- n=1 n n=1 − n ∞ solutely. ThereforeP if (1 pn) converges absolutely then Ppn converges. − n=1 P Q Proof. Let us write p =1+ α . Then, for α < 1, the Taylor series for Log gives n n | n| α2 α3 α4 Log(1 + α ) α = n + n n + n − n − 2 3 − 4 ··· α 2 α 3 α 4 | n| + | n| + | n| + ≤ 2 3 4 ··· α | n| α + α 2 + α 3 + . ≤ 2 | n| | n| | n| ··· If we further assume α 1/2 then the geometric series in the last line has sum at most | n| ≤ 1/2 and so, for α 1/2, Log(1 + α ) α 1 α or, equivalently, | n| ≤ | n − n| ≤ 2 | n| 1 3 1 p Log p 1 p . 2| − n| ≤ n ≤ 2| − n|
The result now follows by the comparison test since clearly for n large enough, we can assume that α 1/2. | n| ≤
33 ∞ ∞ Corollary 20. If α < then (1 + α ) converges. | n| ∞ n n=1 n=1 X Y ∞ ∞ Exercise: Show that α < if and only if (1 + α ) converges. | n| ∞ | n| n=1 n=1 X Y ∞ If either of these above conditions hold then we say that 1 (1+αn) converges absolutely. ∞ ∞ It therefore follows that if 1 (1 + αn) converges absolutelyQ then 1 (1 + αn) converges. This is, of course, because theQ similar result for series is true. Q ∞ To complete the discussion we shall see, by two examples, that the convergence of 1 αn ∞ is neither necessary nor sufficient for the convergence of the product 1 (1 + αk). P −1/2 −1/2 −1 ∞ Example: (a) Let α2n−1 = n and α2n = n + n . ThenQ αn is divergent − n=1 (being the harmonic series). Note that P
1 1 1 1 1+ 1 + =1+ √n − √n n n3/2 ∞ ∞ −3/2 so 1 (1 + αn) = 1 (1 + n ) is convergent. 1 n 2 ∞ Example:Q (b) On theQ other hand if αn =( 1) /n then αn is clearly convergent, whilst − 2 ∞ 2 (1 + αn) is divergent. P Exercise:Q Prove this last statement! Hint: Prove and use the fact that
1 1 1 (1 + α2n−1)(1 + α2n)= 1 1+ < 1 . − √2n 1 √2n − 2n −
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