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 + a

_{n}) is absolutely convergent if and only if ∏ [1 + abs(a

_{n})] 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 - x

^{2}/2 + x

^{3}/3 - ... [See Lemma 1, here]

(2) x - x

^{2}/2 + x

^{3}/3 - ... ≤ x(1 + x + x

^{2}+ ...)

(3) 1 + x + x

^{2}+ ... = 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) ≤ e

^{x}

Proof:

(1) e

^{x}= 1 + x/1! + x

^{2}/2! + ... + x

^{n-1}/(n-1)! [See Lemma 2, here]

(2) Let C = x

^{2}/2! + ... + x

^{n-1}/(n-1)!

(2) If it clear that if x ≥ 0, then:

C ≥ 0

(3) So that we have:

(1 + x) ≤ (1 + x + C) = e

^{x}

QED

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

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

^{x}

(2) So that ln(1 + x) ≤ ln(e

^{x}) = x

QED

Lemma 3: If a

_{n}≥ 0, then ∑ a

_{n}is convergent if and only if ∏ (1 + a

_{n}) is convergent.

Proof:

(1) We know that:

a

_{1}+ a

_{2}+ ... + a

_{n}is less than (1 + a

_{1})(1 + a

_{2})...(1 + a

_{n})

since if you carry out the multiplication, you get:

(1 + a

_{1})(1 + a

_{2})...(1 + a

_{n}) = a

_{1}*1

^{(n-1)}+ a

_{2}*1

^{(n-1)}+ ... + a

_{n}*1

^{(n-1}+ 1 + C

where C = a

_{1}a

_{2}*1*...*1 + a

_{1}a

_{3}*1*...*1 + ... + a

_{1}*a

_{2}*...*a

_{n}

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

1 + a

_{i}≤ e

^{ai}

(3) This gives us that:

(1 + a

_{1})(1 + a

_{2})*...*(1+a

_{n}) ≤ e

^{a1}*e

^{a2}*...*e

^{an}= e

^{a1+a2+...+an}

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

∑ a

_{i}is less than ∏ (1 + a

_{i}) ≤ e

^{∑ai}

(5) So, if ∑ a

_{i}converges, then e

^{∑ai}is finite and ∏ (1 + a

_{i}) converges since it is less than e

^{∑ai}

(6) If ∏ (1 + a

_{i}) converges, then ∑ a

_{i}converges since it is less than ∏ (1 + a

_{i}).

QED

Lemma 4: if ∑ abs(a

_{i}) is convergent, then ∑ abs(log(1 + a

_{i})) is also convergent.

Proof:

(1) Assume ∑ abs(a

_{i}) is convergent.

(2) There exists N such that if n ≥ N, abs(a

_{n}) is less than (1/4) [See Definition 1 here for definition of convergence]

(3) if a

_{i}≥ 0 and i ≥ N, then:

abs[log(1 + a

_{i})] is less than (4/3)abs(a

_{i}) since:

(a) abs[log(1 + a

_{i})] = log(1 + abs[a

_{i}])

Since 1 + a

_{i}≥ 1, we know that log(1 + a

_{i}) ≥ 0.

(b) Using Lemma 1 above, we have:

log(1 + abs[a

_{i}]) ≤ abs(a

_{i})/[1 - abs(a

_{i})] for abs(a

_{i}) is less than 1.

(c) Since abs(a

_{i}) is less than (1/4), we have:

1 - abs(a

_{i}) is greater than 1 - (1/4) which implies that:

1/(1 - abs(a

_{i}) is less than 1/(1 - (1/4)) and further that:

abs(a

_{i})/[1 - abs(a

_{i})] is less than abs(a

_{i})/[1 - (1/4)] = abs(a

_{i})/(3/4) = (4/3)abs(a

_{i})

(4) if a

_{i}is less than 0 and n ≥ N, then:

abs[log(1 + a

_{i})] ≤ (2)abs(a

_{i}) since:

(a) abs[log(1 + a

_{i})] = -log(1 + a

_{i})

Since a

_{i}is less than 0, 1 + a

_{i}is less than 1 and log(1 + a

_{i}) is less than 0.

(b) -log(1 + a

_{i}) = log([1 + a

_{i}]

^{-1}]) = log(1/[1 + a

_{i}]). [See here for review of logarithms if needed and here for review of exponents if needed]

(c) Since 1/(1 + a

_{i}) = ([a

_{i}+ 1] - a

_{i})/(1 + a

_{i}) = 1 - a

_{i}/(1 + a

_{i}), we have:

log(1/[1 + a

_{i}]) = log(1 - a

_{i}/[1 + a

_{i}])

(d) Since a

_{i}is less than 0, we know that:

-a

_{i}/(1 + a

_{i}) = abs(a

_{i})/[1 - abs(a

_{i}) ]

(e) This gives us:

log(1 - a

_{i}/[1 + a

_{i}]) = log(1 + abs[a

_{i}]/[1 - abs(a

_{i})])

(f) Using Lemma 1 above with x = abs(a

_{i})/[1 - abs(a

_{i})], we have:

log(1 + abs[a

_{i}]/[1 - abs(a

_{i})]) ≤ { abs(a

_{i})/[1 - abs(a

_{i})] }/{ 1 - abs(a

_{i})/[1 - abs(a

_{i})] }

since:

abs(a

_{i}) is less than 1/4 (step #2)

1 - abs(a

_{i}) is greater than 1 - 1/4 so that 1/(1 - abs(a

_{i})) is less than 1/(1 - 1/4) = 1/(3/4) = 4/3.

abs(a

_{i})/(1 - abs(a

_{i})) is less than (1/4)*(4/3) = 1/3

This allows us to use Lemma 1 since abs(a

_{i})/[1 - abs(a

_{i})] is less than 1.

(g) We can simplify since:

{ abs(a

_{i})/[1 - abs(a

_{i})] }/{ 1 - abs(a

_{i})/[1 - abs(a

_{i})] } =

={ abs(a

_{i})/[1 - abs(a

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

_{i}]/[1 - abs(a

_{i})] - abs(a

_{i})/[1 - abs(a

_{i})] } =

= { abs(a

_{i})/[1 - abs(a

_{i})] }/{ [1 - 2*abs(a

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

_{i})]} =

= { abs(a

_{i})/[1 - abs(a

_{i})] } * {[1 - abs(a

_{i})]/[1 - 2*abs(a

_{i})]} =

= abs(a

_{i})/[1 - 2*abs(a

_{i})]

(h) Since abs(a

_{i}) is less than (1/4), we have:

abs[log(1 + a

_{i})] ≤ abs(a

_{i})/[1 - 2*abs(a

_{i})] which is less than abs(a

_{i})/[1 - 2*(1/4)] = abs(a

_{i})/(1/2) = 2*abs(a

_{i})

(5) Thus for all values of a

_{i}if i ≥ N, then:

abs[log(1 + a

_{i})] ≤ (2)abs(a

_{i})

(6) The lemma follows directly from this result.

QED

Theorem 5: If ∏[1 + abs(a

_{i})] is convergent, then ∏(1 + a

_{i}) is also convergent.

Proof:

(1) Assume ∏ [1 + abs(a

_{i})] is convergent.

(2) Using Lemma 3 above, we see that ∑ abs(a

_{i}) is also convergent.

(3) Using Lemma 4 above, this also gives us that ∑ abs[log(1 + a

_{i})] is convergent.

(4) Using Theorem 3 here, we see that ∑ log(1 + a

_{i}) is convergent.

(5) Finally from a previous result, we can conclude step #4 that ∏ (1 + a

_{i}) is convergent [See Lemma 3, here]

QED

Theorem 6: If ∏(1 + a

_{i}) is absolutely convergent, then its factors may be rearranged in any order without affecting its limit.

Proof:

(1) Assume ∏(1 + a

_{i}) is absolutely convergent.

(2) So, then ∏[1 + abs(a

_{i})] is convergent. [See Definition 1 above]

(3) Using Lemma 3 above, we see that ∑ abs(a

_{i}) is convergent.

(4) Using Lemma 4 above, we see that ∑ abs[log(1 + a

_{i})] is convergent which means that ∑ log(1 + a

_{i}) is absolutely convergent [See Definition 2, here for definition of absolute convergence]

(5) This means that the terms in ∑ log(1 + a

_{i}) can be rearranged in any order without affecting its sum. [See Theorem 5, here]

(6) This gives us that ∏(1 + a

_{i}) 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 + a

_{i})

(c) So that ∏(1 + a

_{i}) = e

^{log(∏[1 + a}

^{i])}= e

^{∑log(1 + ai)}= e

^{S}

(d) It is clear that for any rearrangement, the limit L = e

^{S}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

Post a Comment