Wednesday, September 27, 2006

Infinite Products

Today, I present some basic ideas about infinite products that I will use later to demonstrate the convergence of the Euler Product Formula.

Definition 1: Convergence of an infinite product

An infinite product ∏ (n ∈ N) cn is said to converge to a limit L ≠ 0 if the partial products Pn approaches L as n approaches where:

Pn = ∏ (0 ≤ m ≤ n) cm

It also makes sense to define divergence for an infinite product.

Definition 2: Divergence of an infinite product

An infinite product ∏ (n ∈ N) cn is said to diverge if either the product does not converge (according to definition 1 above) or if it converges to 0.

So, if any of the factors of an infinite product is 0, then, that infinite product is said to diverge.

Lemma 1: if ∏ cn is convergent, then (lim n → ∞) cn = 1.

Proof:

(1) Assume that an infinite product ∏ cn is convergent.

(2) Let Pn be the partial product of this infinite product such that:

Pn = ∏ (0 ≤ m ≤ n) cm

(3) Then, for each cn of the series:

cn = Pn/(Pn-1)

(4) Since the product is convergent, there exists L such that as n goes to , Pn → L and Pn-1 → L.

(5) It therefore follows that lim (n → ∞) cn = lim (n → ∞) Pn/Pn-1 = L/L = 1.

QED

Corollary 1.1: lim(1 + an)

if an = cn - 1, then:
∏(1 + an) is convergent → lim (n → ∞) an = 0.

Proof:

(1) an = cn - 1

(2) lim (n → ∞) an = lim(n → ∞) cn - 1 = 1 - 1 = 0 [This follows from Lemma 1 above]

QED

Lemma 2: Criteria for Convergence of ∑ an

For any N ≥ 0, if ∑ (n=N, ∞) an is finite, then ∑ (n=0, ∞) an is finite where ai is a real number.

Proof:

(1) Let {an} be a sequence such that each an is a real number.

(2) Assume that ∑ (n=N, ∞) an is finite

(3) Clearly, if N is any natural number, then ∑ (n=0, n less than N) an is finite.

(3) Since, ∑ (n=0, ∞) an = ∑ (n=0, n less than N) an + ∑ (n = N, ∞), it follows that:

∑ (n=0, ∞) an is finite and therefore convergent.

QED

Lemma 3: Criteria for Convergence of ∏(1 + an)

If an ≠ -1 for n ∈ N, then:
∏ (1 + an) converges if and only if ∑ log(1 + an) converges

Proof:

(1) Assume ∑ log(1 + an) converges to S

(2) Let Sn = ∑ (m ≤ n) log(1 + am)

(3) Then:

eSn = ∏ (m ≤ n) (1 + am) since:

elog(x) + log(y) = elog(xy) = xy [See here for review of the properties of Log]

(4) Since lim (n → ∞) Sn = S, this implies that:

lim (n → ∞) esn = eS

(5) Thus, we have shown that:

lim (n → ∞) (1 + an) = eS

(6) Assume that ∏(1 + an) converges

(7) Let ε be a real number greater than 0.

(8) Then, by Lemma 1 above (see here for review of mathematical limits), there exists N such that:

abs(Pn/Pn-1 - 1) is less than ε if m,n ≥ N.

This follows since lim (n → ∞) Pn = L which means:

lim (n → ∞) Pn/Pn-1 = L/L = 1

(9) Let Pn = ∏ (m ≤ N) (1 + an)

(10) If m,n ≥ N, then:

Log(Pn/PN) = Log(Pm/PN * Pn/Pm) = Log(Pm/PN) + Log(Pn/Pm) [See here for properties of Log]

(11) Let m = n-1

(12) Since cn = Pn/Pn-1 = Pn/Pm = an + 1 (see Lemma 1 above for details), we have:

Log(Pn/PN) = Log(Pm/PN) + Log(1 + an)

(13) Since m = n-1 and since we can apply this same argument to m-1, m-2, etc. we get:

Log(Pn/PN) = [Log(Pm-1) + Log(1 + am)] + Log(1 + an) = ∑ (N is less than i ≤ n) Log(1 + ai)

(14) Since lim (n → ∞) Pn = L, it follows that:

lim (n → ∞) Log(Pn/PN) = Log(L/PN)

(15) This shows that:

lim (n → ∞) ∑ (n greater than N) Log(1 + an) = Log(L/PN)

(16) Using Lemma 2 above gives us:

∑ (n ≥ 0) Log(1 + an) is convergent.

QED

Lemma 4: If abs(z) ≤ (1/2), then abs(log(1 + z)) ≤ 2*abs(z)

Proof:

(1) If abs(z) is less than 1, then Log(1 + z) is holomorphic. [Details to be added later, in the mean time see here]

(2) Since Log(1 + z) is holomorphic (see here for definition of holomorphic),

Log(1 + z) = z - z2/2 + z3/3 - ... [See Theorem, here]

(3) Using the Triangle Inequality for Complex Numbers (see Lemma 3, here), we have:

abs(Log(1 + z)) ≤ abs(z) + abs(z2/2) + abs(z3/3) + ...

which is less than

abs(z) + abs(z)2 + abs(z)3 + ...

(4) Now, abs(z) + abs(z)2 + abs(z)3 + ... = abs(z)/[1 - abs(z)] [See Lemma 1, here]

(5) Since abs(z) ≤ (1/2), it follows that:

1/[1 - abs(z)] ≤ 1/[1 - (1/2)] = 1/(1/2) = 2.

(6) So that we have:

abs(Log(1 + z)) ≤ 2*abs(z)

QED

Lemma 5: if ∑ an converges, there exists an natural number N such that for all n ≥ N, an is less than (1/2).

Proof:

(1) If ∑ ai converges, then only a finite number of ai ≥ (1/2).

(a) Assume that there is an infinite number of ai such that ai ≥ (1/2)

(b) If we build a subsequence (see Definition 6, here for definition of subsequence) of just those values of ai where ai ≥ (1/2)

(c) Since there are an infinite number of them, it follows that ∑ ai = ∞

(d) But this is impossible since ∑ (i=1, ∞) ai converges in step #1, so we reject our assumption at step #1a.

(2) If there are only a finite number of ai where ai ≥ (1/2), then we can let N-1 = the last such ai ≥ (1/2).

(3) Then, it follows that for all i ≥ N, ai is less than (1/2).

QED

Lemma 6: Criteria for Convergence using ∑ abs(an)

if an ≠ -1 for n ∈ N, then:

∑ abs(an) is convergent → ∏(1 + an) is convergent

Proof:

(1) Assume that ∑ abs(an) converges such that ∑ abs(an) = L

(2) Then there exists N such that n ≥ N → abs(an) is less than (1/2) [See Lemma 5 above]

(3) Using Lemma 4 above with an, it follows that:

abs(Log(1 + an)) ≤ 2*abs(an)

(4) So that we have:

∑ abs(Log(1 + an)) ≤ ∑ 2*abs(an) = 2*∑ abs(an) = 2*L.

(5) The conclusion follows from applying Lemma 3 above.

QED

References

3 comments :

Kazbich-the-bandit said...

can I ask you for a proof or explanation of why an infinite product is said to diverge when the limit = 0? Thanks

Larry Freeman said...

Hi Kazbich,

You can find Wikipedia's explanation here.

Feel free to let me know if you have additional questions from the Wikipedia article.

Cheers,

-Larry

Kazbich-the-bandit said...

Hi Larry, I have used both your post and the wikipedia article and I think I get the main points. I have two questions, however. In the second part of the proof of convergence criteria for Product (1 + an), you start with the fact that abs(Pn/Pn-1 - 1) is smaller than a no. e greater than zero for m,n greater than N. I am right in thinking that is necessary in order to ensure than the function log (Pn/Pn-1) exists for the interval [N is smaller than m is smaller than n]? Secondly, is this connected to the fact that the sequence of partial products is said to diverge if it converges to zero, since lim (n - infinity) log (Pn/PN) is undefined if Pn converges to zero? Thanks