## Friday, November 03, 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, ..., An is said to converge to L if and only if lim (n → ∞) An = L. [See Definition 1 here for definition of a mathematical limit]

Definition 2: Absolute Convergence

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

Definition 3: Conditional Convergence

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

Lemma 1: Comparison Test

Assume ∑ ai is convergent with all ai ≥ 0.
Assume all bi ≥ 0.

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

Then, ∑ bi is also convergent

Proof:

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

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

QED

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

Proof:

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

(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) ai - ∑ (i=1,n+1) bi =

= ∑ (i=1,n) ai - ∑ (i=1,n) bi + an+1 - bn+1 =

= ∑ (i=1,n) (ai - bi) + (an+1 - bn+1) =

∑ (i=1,n+1) (ai - bi)

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

QED

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

Proof:

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

QED

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

Proof:

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

if ai ≥ 0, then ui = ai

if ai ≤ 0, then ui = 0

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

if ai ≤ 0, then vi = -ai

if ai ≥ 0, then vi = 0

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

(a) all ui, vi ≥ 0

(b) abs(ai) = ui + vi

(c) ai = ui - vi

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

(a) ui ≤ abs(ai)

(b) vi ≤ abs(ai)

(5) If ∑ abs(ai) is convergent, then by Lemma 1 above, ∑ ui and ∑ vi are convergent.

We can use Lemma 1 since:

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

(b) all ui and vi ≥ 0 (see #3a)

(c) N=1, K=1 since all ui ≤ abs(ai) [#4a] and all vi ≤ abs(ai) [#4b]

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

(7) Then, since ai = ui - vi (see #3c), it follows that ∑ ai is also convergent.

QED

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

Proof:

(1) Let bi be defined such that:

if ai ≥ 0, then bi = ai

if ai ≤ 0, then bi = 0

(2) Let ci be defined such that:

if ai ≥ 0, then ci = 0

if ai ≤ 0, then ci = ai

(3) It follows that:

(a) ai = bi + ci

(b) abs(ai) = bi - ci

(4) Let:

An = ∑ (i=1,n) ai

Bn = ∑ (i=1,n) bi

Cn = ∑ (i=1,n) ci

(5) Let An* = ∑ (i=1,n) abs(ai)

(6) Then Bn = (1/2)(An + An*) since:

(a) If we add the two equations ai = bi + ci (#3a) and abs(ai) = bi - ci (#3b), we get:

2*bi = ai + abs(ai)

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

(c) This gives us:

Bn = (1/2)(An + An*)

(7) Cn = (1/2)(An - An*) since:

(a) If we subtract abs(ai) = bi - ci (#3b) from ai = bi + ci (#3a), we get:

2*ci = ai - abs(ai)

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

(c) This gives us:

Cn = (1/2)(An - An*)

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

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

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

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

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

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

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

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

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

For i ≤ n1, let ui = bi

For j ≤ n2, let u(n1+j) = cj

For k ≤ n3, let u(n1+n2+k) = bk

And so on for the rest of the series.

(16) Let Ui = ∑ ui

(17) We can see that:

Un1 is greater than L

Un1 + n2 is less than L

Un1 + n2 + n3 is greater than L

(18) abs(Un1 - L) is less than abs(un1) since:

(a) Un1 is greater than L which is greater than U(n1-1)

(b) abs(L - Un1) is less than abs(U(n1-1) - Un1) = abs(un1)

(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:

Un - L is between U(n1 + n2) - L and Un1 - L since:

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

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

un1 ≥ un greater than u(n1 + n2) [Since u(n1 + n2) is negative and is the least number where ∑ ui is less than L.]

(c) This gives us that:

(Un1 - L) ≥ (Un - L) which is greater than (U(n1+n2) - L)

(21) We can now conclude that abs(Un-L) is less than abs(un1) + abs(u(n1 + n2)) since:

(a) abs(Un -L) ≤ abs(Un1 - L) [#20c]

(b) abs(U(n1+>n2) - L) ≥ 0 [#12]

(c) So, abs(Un -L) ≤ abs(Un1 - L) + abs(U(n1+>n2) - L)

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

abs(Un -L) is less than abs(un1) + abs(u(n1 + n2))

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

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

(23) There exists an integer m such that:

n1 + n2 + ... + nm ≤ n is less than n1 + n2m + nm+1

This follows since ni 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(Un - L) is less than abs(u(n1 + n2 + ... + nm)) + abs((u(n1 + n2 + ... + nm + nm+1))

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

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

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

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

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

(29) It is clear that vi is an infinite sum and ∑ vi = ∑ ui = L.

QED

Theorem 5: Rearrangeable Terms

if:

all ai ≥ 0 and bi is a rearrangement of ai such that for all ai = bo(i) where o(i) is an ordering function on i.

then:

if ∑ ai converges, then ∑ bi does too.

if ∑ ai diverges, then ∑ bi also diverges.

Proof:

(1) Assume that:

b1 = am1

b2 = am2

b3 = am3

and so on

(2) Let Bn = ∑ (i=1,n) bi and An = ∑ (i=1,n) ai

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

(4) So, we have:

Bn = ∑ (i=1,n) ami ≤ ∑ (i=1,p) ai = Ap

(5) For An, let k be the point where Bk ≥ An.

(6) Then, we have:

An = ∑ (i=1,n) ai ≤ ∑ (i=1,k) bi = Bk

(7) Assume that ∑ ai converges to L

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

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

(10) This shows that if ∑ ai converges to L, then ∑ bi also converges to L.

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

QED

References
• Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1989
• James M. Hyslop, , Dover, 2006