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