## 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

## Tuesday, September 26, 2006

### 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 ∑ an 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) = an for all integers n ≥ 1, then the series and the improper integral:

∑ (n=1, ∞) an

∫ (1,∞) f(x)dx

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

Proof:

(1) Let Sn = a1 + a2 + ... + an

(2) If f(n) = an 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 ≤ Sn (see graph below) (3) If we now consider the sum Sn - a1 = a2 + a3 + ... + an, then:

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

Sn - 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 ∑ an diverges.

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

(8) From step #3,

Sn ≤ a1 + ∫(n,1)f(x)dx ≤ a1 + I

(9) So the monotone increasing sequence {Sn} (∞,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 ∑ an = lim (n → ∞) Sn 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/ns = 1 + 1/2s + 1/3s + ... + 1/ns + ...

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/xs satisfies the conditions of the integral test (it is a positive-valued, decreasing, continuous function for x ≥ 1).

(3) ∫ (∞,1) (1/xs)dx = lim (b → ∞) ∫ (b,1) (1/xs)dx = lim (b → ∞) [-1/[(s-1)xs-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)xs-1]] (b,1) = lim(b → ∞) (1/[s-1])[1 - 1/(bs-1)]

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

lim(b → ∞) (1/[s-1])[1 - 1/(bs-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/xs)dx = lim (b → ∞) [1/(1-s)](b1-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

## Sunday, September 24, 2006

### 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)*xi-1

for x ≠ -1

Proof:

(1) Let s = ∑ (i=1,∞) (-1)(i+1)*xi-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)*xi-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