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

^{Sn}= ∏ (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

^{sn}= 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 - z

^{2}/2 + z

^{3}/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(z

^{2}/2) + abs(z

^{3}/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