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, Infinite Series, Dover, 2006
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