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
  

Integral Test for Infinite Series

The Integral Test for Infinite Series is a well known convergence test that uses integrals. The Integral Test, for example, can be used, for example, to show that the harmonic series diverges (that is, does not have a limit).

The content in today's blog is taken from Edwards and Penney, Calculus and Analytic Geometry.

Definition 1: Improper Integral

An Improper Integral is an integral on an interval that is open (does not include its endpoints) or on a closed interval that is unbounded (that is, one of its endpoints is either ∞ or -∞). [See here for a review of open and closed intervals]

Improper integrals are used to analyze infinite series. The improper integral is the limit of a sequence of definite integrals where endpoints get closer and closer to the limit.

Theorem 1: The Integral Test

Let ∑ a_n be a positive-term series (none of the elements are negative) and let f(x) be a function that is a positive valued, decreasing continuous function for x ≥ 1.

If f(n) = a_n for all integers n ≥ 1, then the series and the improper integral:

∑ (n=1, ∞) a_n

∫ (1,∞) f(x)dx

either both converge or diverge. That is, one diverges if and only if the other diverges.

Proof:

(1) Let S_n = a₁ + a₂ + ... + a_n

(2) If f(n) = a_n for all integers n ≥ 1, then it is clear that for the region under y=f(x) from x=1 to x = n+1:

∫ (1,n+1) f(x)dx ≤ S_n (see graph below)

(3) If we now consider the sum S_n - a₁ = a₂ + a₃ + ... + a_n, then:

If f(n) = a_n for all integers n ≥ 1, then it is clear that for the region under y=f(x) from x=1 to x = n:

S_n - a₁ ≤ ∫ (n,1) f(x)dx (see graph below)

(4) Assume that ∫(∞,1) f(x)dx diverges.

Then it goes to +∞ since by assumption f(x) is a positive term series.

(5) lim (n → ∞) ∫ (n+1,1) f(x) = +∞

(6) From step #2, it is clear that lim (n → ∞) Sn = +∞ so that ∑ a_n diverges.

(7) Assume that ∫(∞,1) f(x)dx converges with value = I.

(8) From step #3,

S_n ≤ a₁ + ∫(n,1)f(x)dx ≤ a₁ + I

(9) So the monotone increasing sequence {S_n} (∞,1) is bounded. [See Definition 8, here for definition of monotone increasing sequence, see Definition 4, here for definition of bounded]

(10) This gives that the infinite series ∑ a_n = lim (n → ∞) S_n converges also.

QED

Corollary 1.1: Harmonic series diverges

∑ (n=1, ∞) 1/n = 1 + 1/2 + 1/3 + 1/4 + ...

Proof:

(1) f(x)=1/x is positive, continuous, and decreasing for x ≥ 1.

(2) ∫(∞,1) (1/x)dx = lim (b → ∞) ∫ (b,1) (1/x)dx = lim (b → ∞) [ln x] (b,1) [This follows from the derivative of 1/x, see Lemma 1, here]

(3) lim (b → ∞) [ln x] (b,1) = lim (b → 0) (ln b - ln 1) = + ∞

(4) Therefore, using the Integral Test above, the series diverges.

QED

Corollary 1.2: Zeta function converges if s > 1 but diverges if 0 is less than s ≤ 1.

ζ(s) = ∑ (n=1,∞) 1/n^s = 1 + 1/2^s + 1/3^s + ... + 1/n^s + ...

Proof:

(1) The case where s=1 was proven in Corollary 1.1

(2) If s is greater than 0 and s ≠ 1, then f(x) = 1/x^s satisfies the conditions of the integral test (it is a positive-valued, decreasing, continuous function for x ≥ 1).

(3) ∫ (∞,1) (1/x^s)dx = lim (b → ∞) ∫ (b,1) (1/x^s)dx = lim (b → ∞) [-1/[(s-1)x^s-1]] (b,1) [This follows from the power rule for derivatives (see Lemma 2, here) and the fundamental theorem of calculus]

(4) lim (b → ∞) [-1/[(s-1)x^s-1]] (b,1) = lim(b → ∞) (1/[s-1])[1 - 1/(b^s-1)]

(5) If s is greater than 1, then:

lim(b → ∞) (1/[s-1])[1 - 1/(b^s-1)] = (1/[s-1])(1) = (1/[s-1]) which is less than ∞ so it converges.

(6) If 0 is less than s is less than 1, then:

∫ (∞,1)(1/x^s)dx = lim (b → ∞) [1/(1-s)](b^1-p - 1) = ∞

[This follows from the generalized power rule for derivatives (see Theorem, here) and the fundamental theorem of calculus]

(7) The conclusion follows from the Integral Test above.

QED

References

• Mohamed A. Khamsi, "Improper Integrals: An Introduction", SOSMath.com
• Edwards & Penney, Calculus and Analytic Geometry

Power rule for integrals

Here is another elementary use of the fundamental theorem of calculus.

Lemma: Power rule for integrals

if n ≠ -1, then ∫ axⁿdx = (1/[n+1])axⁿ⁺¹+C

Proof:

(1) d/dx[ (1/[n+1])axⁿ⁺¹+C] = (1/[n+1]a)d/dx[xⁿ⁺¹] + d/dx[C] =
(1/[n+1])(n+1)*axⁿ + 0 = axⁿ

First, we apply Lemma 3, here to split up the sum. Since the derivative is defined in terms of limits (see Definition 1, here), we can move the constants (1/[n+1])*a. The constant rule (see Lemma 1, here) gives us that d/dx[C] = 0. Finally, the power rule for derivatives (see Lemma 2, here) gives us our result.

(2) Since (1/[n+1])axⁿ⁺¹ + C is the antiderivative for axⁿ (see Definition 2, here, for definition of antiderivative), the Fundamental Theorem of Calculus (see Theorem 2, here) gives us our conclusion.

QED

log (1 + x) = ∑ (n=1,∞) [(-1)n+1xn]/n

In today's blog, I use the fundamental theorem of calculus and the basic properties of derivatives to show that:
log (1 + x) = ∑ (n=1,∞) [(-1)ⁿ⁺¹xⁿ]/n

First, we need a lemma:

Lemma 1: 1/(x+1) = ∑ (i=1,∞) (-1)⁽ⁱ⁺¹⁾*x^i-1

for x ≠ -1

Proof:

(1) Let s = ∑ (i=1,∞) (-1)⁽ⁱ⁺¹⁾*x^i-1 which gives us:

s = 1 - x + x² -x³ + ....

(2) xs = x - x² + x³ - x⁴ + ...

(3) Then xs + s = 1

(4) Solving for s gives us:

s(x+1) = 1

so that:

s = 1/(x+1)

QED

Theorem: log (1 + x) = ∑ (n=1,∞) [(-1)ⁿ⁺¹xⁿ]/n

Proof:

(1) From Lemma 1 above;

1/(1+x) = ∑ (i=1,∞) (-1)⁽ⁱ⁺¹⁾*x^i-1

(2) d/dx(ln(1+x)) = 1/(1+x) since:

(a) Let g(x) = 1 + x

(b) Let h(x) = ln(x)

(c) Using the Chain Rule for Derivatives (see Lemma 2, here), we have:

d/dx(ln(1+x)) = d/dx(h(g(x)) = h'(g(x))*g'(x)

where:

h'(g(x)) = ln(1+x) = 1/(1+x) [see Lemma 1, here]

g'(x) = d/dx(1 + x) = d/dx(1) + d/dx(x) = 0 + 1 = 1

so:

d/dx(ln(1+x)) = [1/(1+x)]*1 = 1/(1+x)

(3) Since d/dx(ln(1+x)) = 1/(1+x), we know that ∫(1/(1+x)dx) = ln(1+x)+C [See Theorem 2, here]

(4) If we take the antiderivative of both sides (see Definition 2, here for definition of antiderivative), we have:

ln(1+x) = ∫(1 - x + x² - x³ + ...)

(5) Using the linearity property of integrals (see Lemma, here) gives us:

ln(1+x) = ∫1 - ∫xdx + ∫x²dx - ∫x³dx + ...

(6) Using the power rule for integrals (see Lemma, here) gives us:

ln(1+x) = x - x²/2 + x³/3 -x⁴/4 + ....

QED

Linearity Property of the Integral

The linearity property of the integral refers to the very important property of being able to break integrals apart into sums. That is, ∫(b,a) [f(x)dx + g(x)dx] = ∫ (b,a) f(x)dx + ∫(b,a)g(x)dx.

The content in today's blog is taken from Edwards and Penney's Calculus and Analytic Geometry.

Theorem: Linearity Property of Integrals

If α, β are constants and f(x) and g(x) are continuous functions on [a,b], then:

∫ (b,a) [αf(x) + βg(x)]dx = α∫(b,a)f(x)dx + β∫(b,a)g(x)dx

Proof:

(1) By the Constant Multiple Property (see Lemma 2, here):

∫(b,a) cf(x)dx = c∫(b,a) f(x)dx

(2) Let F(x) be the antiderivative of f(x) and G(x) be the antiderivative of g(x).

(3) d/dx[F(x) + G(x)] = f(x) + g(x) [See Lemma 3, here]

(4) Using the Fundamental Theorem of Calculus (see Thereom 2, here), this gives us:

∫ (b,a) [f(x) + g(x)]dx = [F(x) + G(x)] (b,a)

(5) Using the Evaluation of Integrals (see Theorem 3, here), we have:

[F(x) + G(x)](b,a) = [F(b) + G(b)] - [F(a) + G(a)] = [F(b) - F(a)] + [G(b) - G(a)] =

= [F(x)](b,a) + [G(x)](b,a) = ∫(b,a) f(x)dx + ∫(b,a) g(x)dx.

QED

References

• Edwards & Penney, Calculus and Analytic Geometry