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 + 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 - 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) ≤ e^x

Proof:

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

(2) Let C = x²/2! + ... + 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:

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

since if you carry out the multiplication, you get:

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

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

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

1 + a_i ≤ e^a_i

(3) This gives us that:

(1 + a₁)(1 + a₂)*...*(1+a_n) ≤ e^a₁*e^a₂*...*e^a_n = e^{a₁+a₂+...+a_n}

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

∑ a_i is less than ∏ (1 + a_i) ≤ e^∑a_i

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

(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 + a_i)} = 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

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

Definition 2: Absolute Convergence

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

Definition 3: Conditional Convergence

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

Lemma 1: Comparison Test

Assume ∑ a_i is convergent with all a_i ≥ 0.
Assume all b_i ≥ 0.

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

Then, ∑ b_i is also convergent

Proof:

(1) Since ∑ a_i 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) a_i 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) b_i is less than K * ∑ (i=n+1, n+p) a_i which is less than K*ε/K = ε.

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

QED

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

Proof:

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

(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) a_i - ∑ (i=1,n+1) b_i =

= ∑ (i=1,n) a_i - ∑ (i=1,n) b_i + a_n+1 - b_n+1 =

= ∑ (i=1,n) (a_i - b_i) + (a_n+1 - b_n+1) =

∑ (i=1,n+1) (a_i - b_i)

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

QED

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

Proof:

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

QED

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

Proof:

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

if a_i ≥ 0, then u_i = a_i

if a_i ≤ 0, then u_i = 0

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

if a_i ≤ 0, then v_i = -a_i

if a_i ≥ 0, then v_i = 0

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

(a) all u_i, v_i ≥ 0

(b) abs(a_i) = u_i + v_i

(c) a_i = u_i - v_i

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

(a) u_i ≤ abs(a_i)

(b) v_i ≤ abs(a_i)

(5) If ∑ abs(a_i) is convergent, then by Lemma 1 above, ∑ u_i and ∑ v_i are convergent.

We can use Lemma 1 since:

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

(b) all u_i and v_i ≥ 0 (see #3a)

(c) N=1, K=1 since all u_i ≤ abs(a_i) [#4a] and all v_i ≤ abs(a_i) [#4b]

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

(7) Then, since a_i = u_i - v_i (see #3c), it follows that ∑ a_i is also convergent.

QED

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

Proof:

(1) Let b_i be defined such that:

if a_i ≥ 0, then b_i = a_i

if a_i ≤ 0, then b_i = 0

(2) Let c_i be defined such that:

if a_i ≥ 0, then c_i = 0

if a_i ≤ 0, then c_i = a_i

(3) It follows that:

(a) a_i = b_i + c_i

(b) abs(a_i) = b_i - c_i

(4) Let:

A_n = ∑ (i=1,n) a_i

B_n = ∑ (i=1,n) b_i

C_n = ∑ (i=1,n) c_i

(5) Let A_n* = ∑ (i=1,n) abs(a_i)

(6) Then B_n = (1/2)(A_n + A_n*) since:

(a) If we add the two equations a_i = b_i + c_i (#3a) and abs(a_i) = b_i - c_i (#3b), we get:

2*b_i = a_i + abs(a_i)

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

(c) This gives us:

B_n = (1/2)(A_n + A_n*)

(7) C_n = (1/2)(A_n - A_n*) since:

(a) If we subtract abs(a_i) = b_i - c_i (#3b) from a_i = b_i + c_i (#3a), we get:

2*c_i = a_i - abs(a_i)

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

(c) This gives us:

C_n = (1/2)(A_n - A_n*)

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

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

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

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

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

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

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

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

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

For i ≤ n₁, let u_i = b_i

For j ≤ n₂, let u_(n1+j) = c_j

For k ≤ n₃, let u_(n1+n2+k) = b_k

And so on for the rest of the series.

(16) Let U_i = ∑ u_i

(17) We can see that:

U_n1 is greater than L

U_{n1 + n2} is less than L

U_{n1 + n2 + n3} is greater than L

(18) abs(U_n1 - L) is less than abs(u_n1) since:

(a) U_n1 is greater than L which is greater than U_(n1-1)

(b) abs(L - U_n1) is less than abs(U_(n1-1) - U_n1) = abs(u_n1)

(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:

U_n - L is between U_{(n1 + n2)} - L and U_n1 - L since:

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

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

u_n1 ≥ u_n greater than u_{(n1 + n2)} [Since u_{(n1 + n2)} is negative and is the least number where ∑ u_i is less than L.]

(c) This gives us that:

(U_n1 - L) ≥ (U_n - L) which is greater than (U_(n1+n2) - L)

(21) We can now conclude that abs(U_n-L) is less than abs(u_n1) + abs(u_{(n1 + n2)}) since:

(a) abs(U_n -L) ≤ abs(U_n1 - L) [#20c]

(b) abs(U(_n1+>n2) - L) ≥ 0 [#12]

(c) So, abs(U_n -L) ≤ abs(U_n1 - L) + abs(U(_n1+>n2) - L)

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

abs(U_n -L) is less than abs(u_n1) + abs(u_{(n1 + n2)})

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

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

(23) There exists an integer m such that:

n₁ + n₂ + ... + n_m ≤ n is less than n₁ + n_{2m + nm+1}

This follows since n_i 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(U_n - L) is less than abs(u_{(n1 + n2 + ... + nm)}) + abs((u_{(n1 + n2 + ... + nm + nm+1)})

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

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

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

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

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

(29) It is clear that v_i is an infinite sum and ∑ v_i = ∑ u_i = L.

QED

Theorem 5: Rearrangeable Terms

if:

all a_i ≥ 0 and b_i is a rearrangement of a_i such that for all a_i = b_o(i) where o(i) is an ordering function on i.

then:

if ∑ a_i converges, then ∑ b_i does too.

if ∑ a_i diverges, then ∑ b_i also diverges.

Proof:

(1) Assume that:

b₁ = a_m1

b₂ = a_m2

b₃ = a_m3

and so on

(2) Let B_n = ∑ (i=1,n) b_i and A_n = ∑ (i=1,n) a_i

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

(4) So, we have:

B_n = ∑ (i=1,n) a_mi ≤ ∑ (i=1,p) a_i = A_p

(5) For A_n, let k be the point where B_k ≥ A_n.

(6) Then, we have:

A_n = ∑ (i=1,n) a_i ≤ ∑ (i=1,k) b_i = B_k

(7) Assume that ∑ a_i converges to L

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

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

(10) This shows that if ∑ a_i converges to L, then ∑ b_i also converges to L.

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

QED

References

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

• "List of Trigonometric Identities", Wikipedia
  

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) c_n is said to converge to a limit L ≠ 0 if the partial products P_n approaches L as n approaches ∞ where:

P_n = ∏ (0 ≤ m ≤ n) c_m

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

Definition 2: Divergence of an infinite product

An infinite product ∏ (n ∈ N) c_n 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 ∏ c_n is convergent, then (lim n → ∞) c_n = 1.

Proof:

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

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

P_n = ∏ (0 ≤ m ≤ n) c_m

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

c_n = P_n/(P_n-1)

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

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

QED

Corollary 1.1: lim(1 + a_n)

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

Proof:

(1) a_n = c_n - 1

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

QED

Lemma 2: Criteria for Convergence of ∑ a_n

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

Proof:

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

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

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

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

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

QED

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

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

Proof:

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

(2) Let S_n = ∑ (m ≤ n) log(1 + a_m)

(3) Then:

e^S_n = ∏ (m ≤ n) (1 + a_m) since:

e^{log(x) + log(y)} = e^log(xy) = xy [See here for review of the properties of Log]

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

lim (n → ∞) e^s_n = e^S

(5) Thus, we have shown that:

lim (n → ∞) (1 + a_n) = e^S

(6) Assume that ∏(1 + a_n) 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(P_n/P_n-1 - 1) is less than ε if m,n ≥ N.

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

lim (n → ∞) P_n/P_n-1 = L/L = 1

(9) Let P_n = ∏ (m ≤ N) (1 + a_n)

(10) If m,n ≥ N, then:

Log(P_n/P_N) = Log(P_m/P_N * P_n/P_m) = Log(P_m/P_N) + Log(P_n/P_m) [See here for properties of Log]

(11) Let m = n-1

(12) Since c_n = P_n/P_n-1 = P_n/P_m = a_n + 1 (see Lemma 1 above for details), we have:

Log(P_n/P_N) = Log(P_m/P_N) + Log(1 + a_n)

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

Log(P_n/P_N) = [Log(P_m-1) + Log(1 + a_m)] + Log(1 + a_n) = ∑ (N is less than i ≤ n) Log(1 + a_i)

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

lim (n → ∞) Log(P_n/P_N) = Log(L/P_N)

(15) This shows that:

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

(16) Using Lemma 2 above gives us:

∑ (n ≥ 0) Log(1 + a_n) 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 ∑ a_n converges, there exists an natural number N such that for all n ≥ N, a_n is less than (1/2).

Proof:

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

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

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

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

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

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

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

QED

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

if a_n ≠ -1 for n ∈ N, then:

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

Proof:

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

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

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

abs(Log(1 + a_n)) ≤ 2*abs(a_n)

(4) So that we have:

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

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

QED

References

• James M. Hyslop, Infinite Series, Dover Publications, 2006.
• Complex Logarithms, Wikipedia