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2.6 Infinite Products 2.6. INFINITE PRODUCTS 57 2.6 Infinite Products Infinite product plays important role in mathematics. For example, here are some famous infinite products: 1. Wallis product ∞ π 2 2 4 4 6 6 8 8 4n2 = · · · · · · · · ... = 2 1 3 3 5 5 7 7 9 4n2 − 1 nY=1 2. Euler-Wallis formula ∞ z2 sin πz = πz 1 − . n2 nY=1 3. Euler formula ∞ − ∞ 1 1 1 1 − = ps ns nY=1 n Xn=1 19 where pn is the nth prime, and the right hand side is the Riemann zeta function. Let us prove Euler’s formula: 1 1 1 1 1 1 1 1 1 · 1 · 1 · 1 · 1 ...... =1+ s + s + s + ... 1 − 2s 1 − 3s 1 − 5s 1 − 7s 1 − 11s 2 3 4 In fact, each factor of the left hand side can be written as a geometric series: 1 1 1 1 ..... 1 =1+ s + 2s + 3s + 1 − s pn pn pn pn Multiplying all these series together, we get the series in the right hand side. Here we use a fact that any positive integer can be uniquely written as k1 km ps1 · ... · psm where ps1 , ..., psm are some prime numbers and k1, ..., km are some positive integers. 19The most famous Riemann Hypothesis is sill unsolved. The relationship between Riemann zeta function and the number theory is clear. 58 CHAPTER 2. COMPLEX-VALUED FUNCTIONS ∞ ∞ Convergence of Infinite product zn Given a sequence {zn}n , an infinite product ∞ n=0 =0 n=0 zn is called convergent if the followingQ are satisfied: Q (i) ∃n0 > 0 such that an =06 , ∀n ≥ n0. n0+m ∞ (ii) { n=n0 }m=0 converges to a non-zero complex number. Q 20 ∞ Theorem 2.6.1 (i) If n=0 zn converges, then limn→∞ zn =1. Q ∞ n2 (ii) n=0 zn converges ⇐⇒ ∀ǫ> 0, ∃n0 such that n=n1 zn − 1 < ǫ, ∀n2 ≥ n1 ≥ n0. Q Q Remark: 1. (i) is true because Then ∀m ≥ n0, m zn a z = n=n0 → =1, as m →∞. m Qm−1 a n=n0 zn Q 2. (i) is analogue of the fact for a series: if n an converges, then an → 0. P 3. (ii) is analogue of the fact for a series: n an converges if and only if ∀ǫ > 0, there exists N > 0 such that P |am − an| < ǫ, ∀m>n>N. ∞ Convergence of Infinite Product n=0(1+zn) Recall the complex-valued function log z is defined on ∆(1; 1) while log(1 + z) is well-defined on the unit disk ∆(1), it is convenient ∞ Q ∞ to deal with n=0(1 + zn) instead of n=0 zn. Q Q ∞ Theorem 2.6.2 n=0(1 + zn) converges if and only if ∃n0 such that ∃n0 such that Q ∞ |zn| < 1, ∀n ≥ n0 and log(1 + zn) converges. nX=n0 20 ∞ ∞ ∞ Think this way: n=0 zn = C ⇐⇒ log n=0 zn = log C, i.e., n log zn = log C =⇒ log zn → 0 =⇒ zn → 1. Q Q P 2.6. INFINITE PRODUCTS 59 Proof: (=⇒) By above theorem, ∃n0 such that n 1 1 |z | < , (1 + z ) − 1 < , ∀n ≥ n n 2 j 2 0 jY=n0 ∞ because n=0(1 + zn) converges. Q =⇒ Since log w is a continuous function and has property: limn log wn = log lim wn, we have ∞ log n (1 + z ) → log (1 + z ) , j=n0 j j=n0 j Q k Q n j=n0 log(1 + zj) ∀n → +∞. P ∞ =⇒ n=n0 log(1 + zn) converges. P (⇐=) Because ew is a continuous function, n n n ∞ log Qj=n (1+zj ) Pj=n log(1+zj ) Pj=n log(1+zj ) (1 + zn)= e 0 = e 0 → e 0 =06 . jY=n0 log w ∞ Here we used the identity e = w. =⇒ j=0(1 + zn) converges. Q ∞ 1 [Example] Show: n=1(1 + n2 ) converges. Q ∞ 1 1 Proof: By Theorem above, we need to show: n=1 log(1 + n2 ) converges. Since n n2 converges and P P 1 log(1 + 2 ) log(1 + t) 1 lim n = lim = lim =1, n→∞ 1 t→0 t→0 n2 t 1+ t ∞ 1 we conclude n=1 log(1+ n2 ) converges by applying the Comparison Convergence Test. P.
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