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(ai)/[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, Infinite Series, Dover, 2006
4 comments :
lemma 1 is not true for x>1: the series 1+x+x^2+... has a singularity in x=1, and so cannot be extended from 0 to infinity
Thanks very much for your comment.
I agree with you that I should make it clear that abs(x) ≠ 1.
I've updated the lemma.
-Larry
Lemma 1 is wrong, lemma 4 is wrong for your a,i greater or equal to zero.
Lemma 1 is true for |x|<1. Thus ur lemma 4 is correct for the terms
-1 < a,i <0.
Denote a,i as a subscript i
For a,i>=0, use this ln (1+|a,i|) = ln (1+a,i) <= a,i = |a,i|
which follows easily from ln (1+x)<= x, which follows from e^x>= 1+x
Anonymous,
Thanks very much for noticing the typos. I have updated the lemmas that you mention.
In reviewing it, I realize that these lemmas are far from being as clear as they should be.
When I get a chance, I plan to simplify them. Please continue to let me know if you notice any more inaccuracies.
Thanks very much,
-Larry
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