## 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

## Friday, November 03, 2006

### Infinite Sums

When reasoning about infinite sums, it is necessary to have a good understanding of basic properties.

If one treats infinite sums as finite sums, then contradictions arise. For example, if one is not careful, is it possible to argue that ∞ = -1.

Here's the argument:

(1) ∑ (i=0, ∞) 2i = 1 + 2 + 4 + 8 + ... = ∞

(2) Let T = ∑ (i=1, ∞) = ∞

(3) 2*T = 2 + 4 + 8 + ...

(4) 2*T = T - 1

(5) Then, T = -1

The fallacy here comes in reasoning about a divergent infinite sum. For these reasons, it is important to prove that infinite sums are convergent and further, in this blog, I will show that it is important to note whether a convergent infinite sum is absolutely convergent or conditionally convergent.

Definition 1: Convergence

The sequence A1, A2, ..., An is said to converge to L if and only if lim (n → ∞) An = L. [See Definition 1 here for definition of a mathematical limit]

Definition 2: Absolute Convergence

A sum ∑ an is absolutely convergent if and only if ∑ abs(an) is convergent.

Definition 3: Conditional Convergence

A sum ∑ an is conditionally convergent if and only if ∑ abs(an) is not convergent while ∑ an is convergent.

Lemma 1: Comparison Test

Assume ∑ ai is convergent with all ai ≥ 0.
Assume all bi ≥ 0.

If there exists K, N such that for all n greater than N, bn is less than K*an

Then, ∑ bi is also convergent

Proof:

(1) Since ∑ ai is convergent, then for ε/K, there exists an integer N1 such that for all values of n greater than N1 and for any positive integer p:

∑ (i=n+1, n+p) ai is less than ε/K. [See Definition 1 above]

(2) Let N2 = max(N,N1)

(3) So, that, when n is greater than N2, we have:

∑ (i=n+1, n+p) bi is less than K * ∑ (i=n+1, n+p) ai which is less than K*ε/K = ε.

(4) Since this inequality holds for all positive integral values of p, it follows that ∑ bi is also convergent.

QED

Lemma 2: if ∑ ai is convergent with limit A and ∑ bi is convergent with limit B, then ∑ (ai - bi) is convergent with limit A - B

Proof:

(1) This follows from the fact that for any value of n, ∑ (i=1,n) ai - ∑ (i=1,n) bi = ∑ (i=1,n) (ai - bi)

(2) For n=1, this is obvious since:

a1 - b1 = (a1 - b1)

(3) We assume that it is true up to n where n ≥ 1.

(4) ∑ (i=1,n+1) ai - ∑ (i=1,n+1) bi =

= ∑ (i=1,n) ai - ∑ (i=1,n) bi + an+1 - bn+1 =

= ∑ (i=1,n) (ai - bi) + (an+1 - bn+1) =

∑ (i=1,n+1) (ai - bi)

(5) The result follows as we let n approach infinity.

QED

Corollary 2.1: if ∑ ai is convergent with limit A and ∑ bi is convergent with limit B, then ∑ (ai + bi) is convergent with limit A + B

Proof:

This follows directly from Lemma 2 if we apply Lemma 2 to ∑ -bi which is convergent to limit -B.

QED

Theorem 3: if ∑ abs(ai) is convergent, then ∑ ai is convergent.

Proof:

(1) Let ui be a sequence of terms such that:

if ai ≥ 0, then ui = ai

if ai ≤ 0, then ui = 0

(2) Let vi be a sequence of terms such that:

if ai ≤ 0, then vi = -ai

if ai ≥ 0, then vi = 0

(3) By the above definitions, we see that:

(a) all ui, vi ≥ 0

(b) abs(ai) = ui + vi

(c) ai = ui - vi

(4) From #3b, it is clear that:

(a) ui ≤ abs(ai)

(b) vi ≤ abs(ai)

(5) If ∑ abs(ai) is convergent, then by Lemma 1 above, ∑ ui and ∑ vi are convergent.

We can use Lemma 1 since:

(a) ∑ abs(ai) is convergent and all abs(ai) ≥ 0.

(b) all ui and vi ≥ 0 (see #3a)

(c) N=1, K=1 since all ui ≤ abs(ai) [#4a] and all vi ≤ abs(ai) [#4b]

(6) Using Lemma 2 above, we can conclude that ∑ (ui - vi) is also convergent.

(7) Then, since ai = ui - vi (see #3c), it follows that ∑ ai is also convergent.

QED

Theorem 4: if ∑ ai is conditionally convergent, then for any value L, it is possible to reorder ai and create an infinite sum ∑ ao(i) such that ∑ ao(i) = L and o(i) is an ordering function on i.

Proof:

(1) Let bi be defined such that:

if ai ≥ 0, then bi = ai

if ai ≤ 0, then bi = 0

(2) Let ci be defined such that:

if ai ≥ 0, then ci = 0

if ai ≤ 0, then ci = ai

(3) It follows that:

(a) ai = bi + ci

(b) abs(ai) = bi - ci

(4) Let:

An = ∑ (i=1,n) ai

Bn = ∑ (i=1,n) bi

Cn = ∑ (i=1,n) ci

(5) Let An* = ∑ (i=1,n) abs(ai)

(6) Then Bn = (1/2)(An + An*) since:

(a) If we add the two equations ai = bi + ci (#3a) and abs(ai) = bi - ci (#3b), we get:

2*bi = ai + abs(ai)

(b) Since this is true for each term, we get: 2*Bn = An + An*

(c) This gives us:

Bn = (1/2)(An + An*)

(7) Cn = (1/2)(An - An*) since:

(a) If we subtract abs(ai) = bi - ci (#3b) from ai = bi + ci (#3a), we get:

2*ci = ai - abs(ai)

(b) Since this is true for each term, we get: 2*Cn = An - An*

(c) This gives us:

Cn = (1/2)(An - An*)

(8) Since ∑ ai is conditionally convergent (see Definition 3 above), we know that:

∑ ai is convergent but ∑ abs(ai) is divergent.

(9) This means that ∑ bi is a divergent series of nonnegative terms. [See step #6]

(10) This also means that ∑ ci is a divergent series of nonpositive terms. [See step #7]

(11) Let n1 be the least integer such that ∑ (i=1,n1) bi is greater than L

(12) Let n2 be the least integer such that ∑ (i=1,n1)bi + ∑ (j=1,n2) cj is less than L.

(13) Let n3 be the least integer such that ∑ (i=1,n1 + n3)bi + ∑ (j=1,n2)cj is greater than L.

(14) We can likewise define all ni in a similar manner.

(15) We can now define a new series ∑ ui such that:

For i ≤ n1, let ui = bi

For j ≤ n2, let u(n1+j) = cj

For k ≤ n3, let u(n1+n2+k) = bk

And so on for the rest of the series.

(16) Let Ui = ∑ ui

(17) We can see that:

Un1 is greater than L

Un1 + n2 is less than L

Un1 + n2 + n3 is greater than L

(18) abs(Un1 - L) is less than abs(un1) since:

(a) Un1 is greater than L which is greater than U(n1-1)

(b) abs(L - Un1) is less than abs(U(n1-1) - Un1) = abs(un1)

(19) Likewise, abs(U(n1+n2) - L) is less than abs(u(n1+n2)) since:

(a) U(n1+n2) is less than L which is less than U(n1+n2-1)

(b) abs(L - U(n1+n2)) is less than abs(U(n1+n2-1) - U(n1+n2)) = abs(u(n1+n2))

(20) If n is in between n1 and (n1 + n2), then:

Un - L is between U(n1 + n2) - L and Un1 - L since:

(a) Assume n1 less than n less than n1 + n2

(b) Since ci consists only of nonpositive numbers, we know that:

un1 ≥ un greater than u(n1 + n2) [Since u(n1 + n2) is negative and is the least number where ∑ ui is less than L.]

(c) This gives us that:

(Un1 - L) ≥ (Un - L) which is greater than (U(n1+n2) - L)

(21) We can now conclude that abs(Un-L) is less than abs(un1) + abs(u(n1 + n2)) since:

(a) abs(Un -L) ≤ abs(Un1 - L) [#20c]

(b) abs(U(n1+>n2) - L) ≥ 0 [#12]

(c) So, abs(Un -L) ≤ abs(Un1 - L) + abs(U(n1+>n2) - L)

(d) Using step #18 and step #19, we can conclude that:

abs(Un -L) is less than abs(un1) + abs(u(n1 + n2))

(22) Since ∑ ai is convergent, there exists N such that:

if i is greater than N, abs(ai) is less than (1/2)ε

(23) There exists an integer m such that:

n1 + n2 + ... + nm ≤ n is less than n1 + n2m + nm+1

This follows since ni is a monotonic increasing number as given by the definition in steps #11 thru #14

(24) We can use the same reasoning as in step #21 to conclude that:

abs(Un - L) is less than abs(u(n1 + n2 + ... + nm)) + abs((u(n1 + n2 + ... + nm + nm+1))

(25) Further, we note that all abs(ai) is less than (1/2ε), we can conclude that:

abs(Un - L) is less than (1/2ε) + (1/2ε) = ε

(26) Hence, we have proved that L is also the limit for ∑ ui [See Definition 1 of Convergence above]

(27) In our building of ∑ ui, there is a strong possibility of zeros being added. (See step #15)

(28) Let vi be defined such that it has the same order as ui but only includes values where ui ≠ 0.

(29) It is clear that vi is an infinite sum and ∑ vi = ∑ ui = L.

QED

Theorem 5: Rearrangeable Terms

if:

all ai ≥ 0 and bi is a rearrangement of ai such that for all ai = bo(i) where o(i) is an ordering function on i.

then:

if ∑ ai converges, then ∑ bi does too.

if ∑ ai diverges, then ∑ bi also diverges.

Proof:

(1) Assume that:

b1 = am1

b2 = am2

b3 = am3

and so on

(2) Let Bn = ∑ (i=1,n) bi and An = ∑ (i=1,n) ai

(3) For a set m1, m2, ... , mn, let p = the largest of the integers.

(4) So, we have:

Bn = ∑ (i=1,n) ami ≤ ∑ (i=1,p) ai = Ap

(5) For An, let k be the point where Bk ≥ An.

(6) Then, we have:

An = ∑ (i=1,n) ai ≤ ∑ (i=1,k) bi = Bk

(7) Assume that ∑ ai converges to L

(8) Then step #4, tells us that as n goes toward infinity, Bn ≤ L

(9) And step #6 tells us that as n goes toward infinity, Bn ≥ L

(10) This shows that if ∑ ai converges to L, then ∑ bi also converges to L.

(11) If ∑ ai is divergent, then step #6 tells that ∑ bi must also diverge.

QED

References
• Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1989
• James M. Hyslop, , Dover, 2006

## Wednesday, October 25, 2006

### cos(z) = sin(π/2 + z) and other identities

In today's blog, I go through a through a set of addition properties of sin, cos, tan, and cot.

Lemma 1: cos(z) = sin(z + π/2)

Proof:

(1) sin(π/2 + z) = sin(π/2)cos(z) + sin(z)cos(π/2)

(2) sin(π/2) = 1, cos(π/2)=0 [See Property 2, Property 7, here, see here for explanation on radians if needed]

(3) This gives us:

sin(π/2 + z) = cos(z)*1 + sin(z)*0.

QED

Corollary 1.1: cos(π/2 + x) = -sin(x)

Proof:

(1) -sin(x) = -sin(x + π/2 - π/2) = sin(-x + π/2 - π/2) [See Property 4, here]

(2) sin(π/2 + [-x - π/2]) = cos(-x -π/2) [From Lemma 1 above]

(3) cos(-x -π/2) = cos(x + π/2) [See Property 9, here]

QED

Corollary 1.2: sin(x + π) = -sin(x)

Proof:

(1) sin(x + π) = sin(x + π/2 + π/2) = cos(x + π/2) [See Lemma 1 above]

(2) cos(x + π/2) = -sin(x) [See Corollary 1.1 above]

QED

Corollary 1.3: cos(x + π) = -cos(x)

Proof:

(1) cos(x + π) = cos(x + π/2 + π/2) = -sin(x + π/2) [See Corollary 1.1 above]

(2) -sin(x + π/2) = -cos(x) [See Lemma 1 above]

QED

Corollary 1.4: tan(x + π) = tan(x)

Proof:

(1) tan(x + π) = sin(x + π)/cos(x + π) = -sin(x)/[-cos(x)] = sin(x)/cos(x) = tan(x) [See Corollary 1.2 and 1.3 above]

QED

Corollary 1.5: cot(x + π) = cot(x)

Proof:

(1) cot(x + π) = 1/tan(x + π) = 1/tan(x) = cot(x) [See Corollary 1.4 above]

QED

Corollary 1.6: tan(x + π/2) = -cot(x)

Proof:

(1) tan(x + π/2) = sin(x + π/2)/cos(x + π/2) = cos(x)/[-sin(x)] = -cot(x) [See Lemma 1 and Corollary 1.1 above]

QED

Corollary 1.7: sin(x) = cos(π/2 - x)

Proof:

(1) cos(π/2 - x) = cos(π/2)cos(-x) - sin(π/2)sin(-x) [See Theorem 2, here]

(2) cos(-x) = x [See Property 9, here]

(3) sin(-x) = -sin(x) [See Property 4, here]

(4) cos(π/2)cos(-x) - sin(π/2)sin(-x) = cos(90°)cos(x) + sin(90°)sin(x)

(5) cos(90°) = 0 [See Property 7, here]

(6) sin(90°) = 1 [See Property 2, here]

(7) cos(π/2 - x) = 0*cos(x) + 1*sin(x) = sin(x)

QED

Corollary 1.8: cos(x) = sin(π/2 - x)

Proof:

(1) sin(x) = cos(π/2 - x) [See Corollary 1.7 above]

(2) Let x = π/2 - z

(3) sin(π/2 - z) = cos(π/2 - [π/2 - z]) = cos(z)

QED

Lemma 2: sin(2x) = 2sin(x)cos(x)

Proof:

(1) sin(x + x) = sin(x)cos(x) + cos(x)sin(x) [See Theorem 1, here]

(2) sin(2x) = 2sin(x)cos(x)

QED

Lemma 3: cos(2x) = cos2(x) - sin2(x)

Proof:

(1) cos(x + x) = cos(x)*cos(x) - sin(x)*sin(x) [See Theorem 2, here]

(2) cos(2x) = cos2(x) - sin2(x)

QED

Corollary 3.1: cos(2x) = 2cos2(x) - 1 = 1 - 2sin2(x)

Proof:

(1) cos(2x) = cos2(x) - sin2(x) [See Lemma 3 above]

(2) cos2(x) + sin2(x) = 1. [See Corollary 2, here]

(3) From step #2, we have:

cos2(x) - 1 = -sin2(x)

and also

cos2(x) = 1 - sin2(x)

(3) Combining step #3 and step #1 gives us:

cos(2x) = (1 - sin2(x)) - sin2(x) = 1 - 2sin2(x)

cos(2x) = cos2(x) + (cos2 - 1) = 2cos2(x) - 1

QED

Corollary 3.2: sin2(x) = [1 - cos(2x)]/2

Proof:

(1) cos(2x) = 1 - 2sin2(x) [See Corollary 3.1 above]

(2) 2sin2(x) = 1 - cos(2x)

(3) So that we have:

sin2(x) = [1 - cos(2x)]/2

QED

Corollary 3.3: cos2(x) = [1 + cos(2x)]/2

Proof:

(1) cos(2x) = 2cos2(x) - 1 [See Corollary 3.1 above]

(2) 2cos2(x) = cos(2x) + 1

(3) cos2(x) = [cos(2x) + 1]/2

QED

Corollary 3.4: cos(x/2) = ± √(1/2)[1 + cos(x)]

Proof:

(1) cos2(x/2) = [1 + cos(x)]/2. [From Corollary 3.3 above]

(2) cos(x/2) = ± √(1/2)[1 + cos(x)]

QED

Corollary 3.5: sin(x/2) = ± √(1/2)[1 - cos(x)]

Proof:

(1) sin2(x/2) = [1 - cos(x)]/2. [From Corollary 3.2 above]

(2) sin(x/2) = ± √(1/2)[1 - cos(x)]

QED

Corollary 3.6: cot(2x) = (1/2)[cot(x) - tan(x)]

Proof:

(1) cot(2x) = cos(2x)/sin(2x) = [cos2(x) - sin2(x)]/(2sin(x)cos(x)) =

= (1/2)[cos(x)/sin(x) - sin(x)/cos(x)] = (1/2)[cot(x) - tan(x)]

QED

References

## 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