## Thursday, November 09, 2006

### Absolute Convergence for Infinite Products

In a previous blog, I spoke about absolute convergence in relation to an infinite sum. Absolute convergence is also an important concept for infinite products. Like infinite sums, infinite products that possess absolute convergence can be rearranged in any order and still have the same limit.

The content in today's blog is taken straight from James M. Hyslop's Infinite Series (see reference below).

Definition 1: absolutely convergent infinite product

An infinite product ∏ (1 + an) is absolutely convergent if and only if ∏ [1 + abs(an)] is convergent.

[See Definition 1, here for review of convergence if needed.]

Lemma 1: log (1 + x) ≤ x/(1 - x)

for abs(x) is less than 1.

Proof:

(1) log(1 + x) = x - x2/2 + x3/3 - ... [See Lemma 1, here]

(2) x - x2/2 + x3/3 - ... ≤ x(1 + x + x2 + ...)

(3) 1 + x + x2 + ... = 1/(1 -x) [See Lemma 1, here]

(4) So log(1 + x) ≤ x*[1/(1-x)] = x/(1 - x)

QED

Lemma 2: if x ≥ 0, then (1 + x) ≤ ex

Proof:

(1) ex = 1 + x/1! + x2/2! + ... + xn-1/(n-1)! [See Lemma 2, here]

(2) Let C = x2/2! + ... + xn-1/(n-1)!

(2) If it clear that if x ≥ 0, then:

C ≥ 0

(3) So that we have:

(1 + x) ≤ (1 + x + C) = ex

QED

Corollary 2.1: If x ≥ 0, then ln(1 + x) ≤ x

(1) From Lemma 2, if x ≥ 0, then (1 + x) ≤ ex

(2) So that ln(1 + x) ≤ ln(ex) = x

QED

Lemma 3: If an ≥ 0, then ∑ an is convergent if and only if ∏ (1 + an) is convergent.

Proof:

(1) We know that:

a1 + a2 + ... + an is less than (1 + a1)(1 + a2)...(1 + an)

since if you carry out the multiplication, you get:

(1 + a1)(1 + a2)...(1 + an) = a1*1(n-1) + a2*1(n-1) + ... + an*1(n-1 + 1 + C

where C = a1a2*1*...*1 + a1a3*1*...*1 + ... + a1*a2*...*an

(2) Using Lemma 2 above, we know that:

1 + ai ≤ eai

(3) This gives us that:

(1 + a1)(1 + a2)*...*(1+an) ≤ ea1*ea2*...*ean = ea1+a2+...+an

(4) This means that for all i, we have:

∑ ai is less than ∏ (1 + ai) ≤ e∑ai

(5) So, if ∑ ai converges, then e∑ai is finite and ∏ (1 + ai) converges since it is less than e∑ai

(6) If ∏ (1 + ai) converges, then ∑ ai converges since it is less than ∏ (1 + ai).

QED

Lemma 4: if ∑ abs(ai) is convergent, then ∑ abs(log(1 + ai)) is also convergent.

Proof:

(1) Assume ∑ abs(ai) is convergent.

(2) There exists N such that if n ≥ N, abs(an) is less than (1/4) [See Definition 1 here for definition of convergence]

(3) if ai ≥ 0 and i ≥ N, then:

abs[log(1 + ai)] is less than (4/3)abs(ai) since:

(a) abs[log(1 + ai)] = log(1 + abs[ai])

Since 1 + ai ≥ 1, we know that log(1 + ai) ≥ 0.

(b) Using Lemma 1 above, we have:

log(1 + abs[ai]) ≤ abs(ai)/[1 - abs(ai)] for abs(ai) is less than 1.

(c) Since abs(ai) is less than (1/4), we have:

1 - abs(ai) is greater than 1 - (1/4) which implies that:

1/(1 - abs(ai) is less than 1/(1 - (1/4)) and further that:

abs(ai)/[1 - abs(ai)] is less than abs(ai)/[1 - (1/4)] = abs(ai)/(3/4) = (4/3)abs(ai)

(4) if ai is less than 0 and n ≥ N, then:

abs[log(1 + ai)] ≤ (2)abs(ai) since:

(a) abs[log(1 + ai)] = -log(1 + ai)

Since ai is less than 0, 1 + ai is less than 1 and log(1 + ai) is less than 0.

(b) -log(1 + ai) = log([1 + ai]-1]) = log(1/[1 + ai]). [See here for review of logarithms if needed and here for review of exponents if needed]

(c) Since 1/(1 + ai) = ([ai + 1] - ai)/(1 + ai) = 1 - ai/(1 + ai), we have:

log(1/[1 + ai]) = log(1 - ai/[1 + ai])

(d) Since ai is less than 0, we know that:

-ai/(1 + ai) = abs(ai)/[1 - abs(ai) ]

(e) This gives us:

log(1 - ai/[1 + ai]) = log(1 + abs[ai]/[1 - abs(ai)])

(f) Using Lemma 1 above with x = abs(ai)/[1 - abs(ai)], we have:

log(1 + abs[ai]/[1 - abs(ai)]) ≤ { abs(ai)/[1 - abs(ai)] }/{ 1 - abs(ai)/[1 - abs(ai)] }

since:

abs(ai) is less than 1/4 (step #2)

1 - abs(ai) is greater than 1 - 1/4 so that 1/(1 - abs(ai)) is less than 1/(1 - 1/4) = 1/(3/4) = 4/3.

abs(ai)/(1 - abs(ai)) is less than (1/4)*(4/3) = 1/3

This allows us to use Lemma 1 since abs(ai)/[1 - abs(ai)] is less than 1.

(g) We can simplify since:

{ abs(ai)/[1 - abs(ai)] }/{ 1 - abs(ai)/[1 - abs(ai)] } =

={ abs(a
i)/[1 - abs(ai)] }/{ ([1-abs(ai]/[1 - abs(ai)] - abs(ai)/[1 - abs(ai)] } =

= { abs(ai)/[1 - abs(ai)] }/{ [1 - 2*abs(ai)]/[1 - abs(ai)]} =

= { abs(ai)/[1 - abs(ai)] } * {[1 - abs(ai)]/[1 - 2*abs(ai)]} =

= abs(ai)/[1 - 2*abs(ai)]

(h) Since abs(ai) is less than (1/4), we have:

abs[log(1 + ai)] ≤ abs(ai)/[1 - 2*abs(ai)] which is less than abs(ai)/[1 - 2*(1/4)] = abs(ai)/(1/2) = 2*abs(ai)

(5) Thus for all values of ai if i ≥ N, then:

abs[log(1 + ai)] ≤ (2)abs(ai)

(6) The lemma follows directly from this result.

QED

Theorem 5: If ∏[1 + abs(ai)] is convergent, then ∏(1 + ai) is also convergent.

Proof:

(1) Assume ∏ [1 + abs(ai)] is convergent.

(2) Using Lemma 3 above, we see that ∑ abs(ai) is also convergent.

(3) Using Lemma 4 above, this also gives us that ∑ abs[log(1 + ai)] is convergent.

(4) Using Theorem 3 here, we see that ∑ log(1 + ai) is convergent.

(5) Finally from a previous result, we can conclude step #4 that ∏ (1 + ai) is convergent [See Lemma 3, here]

QED

Theorem 6: If ∏(1 + ai) is absolutely convergent, then its factors may be rearranged in any order without affecting its limit.

Proof:

(1) Assume ∏(1 + ai) is absolutely convergent.

(2) So, then ∏[1 + abs(ai)] is convergent. [See Definition 1 above]

(3) Using Lemma 3 above, we see that ∑ abs(ai) is convergent.

(4) Using Lemma 4 above, we see that ∑ abs[log(1 + ai)] is convergent which means that ∑ log(1 + ai) is absolutely convergent [See Definition 2, here for definition of absolute convergence]

(5) This means that the terms in ∑ log(1 + ai) can be rearranged in any order without affecting its sum. [See Theorem 5, here]

(6) This gives us that ∏(1 + ai) can also be rearranged without affecting its product since:

(a) We know that any new arrangement of the product is convergent by Lemma 3, here.

(b) Let S = ∑ log(1 + ai)

(c) So that ∏(1 + ai) = elog(∏[1 + ai]) = e∑log(1 + ai) = eS

(d) It is clear that for any rearrangement, the limit L = eS which is the same regardless of all the rearrangement so L for each arrangement must likewise be the same.

QED

References
• James M. Hyslop, , Dover, 2006