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, ∞) 2

^{i}= 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 A

_{1}, A

_{2}, ..., A

_{n}is said to converge to L if and only if lim (n → ∞) A

_{n}= L. [See Definition 1 here for definition of a mathematical limit]

Definition 2: Absolute Convergence

A sum ∑ a

_{n}is absolutely convergent if and only if ∑ abs(a

_{n}) is convergent.

Definition 3: Conditional Convergence

A sum ∑ a

_{n}is conditionally convergent if and only if ∑ abs(a

_{n}) is not convergent while ∑ a

_{n}is convergent.

Lemma 1: Comparison Test

Assume ∑ a

_{i}is convergent with all a

_{i}≥ 0.

Assume all b

_{i}≥ 0.

If there exists K, N such that for all n greater than N, b

_{n}is less than K*a

_{n}

Then, ∑ b

_{i}is also convergent

Proof:

(1) Since ∑ a

_{i}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) a

_{i}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) b

_{i}is less than K * ∑ (i=n+1, n+p) a

_{i}which is less than K*ε/K = ε.

(4) Since this inequality holds for all positive integral values of p, it follows that ∑ b

_{i}is also convergent.

QED

Lemma 2: if ∑ a

_{i}is convergent with limit A and ∑ b

_{i}is convergent with limit B, then ∑ (a

_{i}- b

_{i}) is convergent with limit A - B

Proof:

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

_{i}- ∑ (i=1,n) b

_{i}= ∑ (i=1,n) (a

_{i}- b

_{i})

(2) For n=1, this is obvious since:

a

_{1}- b

_{1}= (a

_{1}- b

_{1})

(3) We assume that it is true up to n where n ≥ 1.

(4) ∑ (i=1,n+1) a

_{i}- ∑ (i=1,n+1) b

_{i}=

= ∑ (i=1,n) a

_{i}- ∑ (i=1,n) b

_{i}+ a

_{n+1}- b

_{n+1}=

= ∑ (i=1,n) (a

_{i}- b

_{i}) + (a

_{n+1}- b

_{n+1}) =

∑ (i=1,n+1) (a

_{i}- b

_{i})

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

QED

Corollary 2.1: if ∑ a

_{i}is convergent with limit A and ∑ b

_{i}is convergent with limit B, then ∑ (a

_{i}+ b

_{i}) is convergent with limit A + B

Proof:

This follows directly from Lemma 2 if we apply Lemma 2 to ∑ -b

_{i}which is convergent to limit -B.

QED

Theorem 3: if ∑ abs(a

_{i}) is convergent, then ∑ a

_{i}is convergent.

Proof:

(1) Let u

_{i}be a sequence of terms such that:

if a

_{i}≥ 0, then u

_{i}= a

_{i}

if a

_{i}≤ 0, then u

_{i}= 0

(2) Let v

_{i}be a sequence of terms such that:

if a

_{i}≤ 0, then v

_{i}= -a

_{i}

if a

_{i}≥ 0, then v

_{i}= 0

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

(a) all u

_{i}, v

_{i}≥ 0

(b) abs(a

_{i}) = u

_{i}+ v

_{i}

(c) a

_{i}= u

_{i}- v

_{i}

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

(a) u

_{i}≤ abs(a

_{i})

(b) v

_{i}≤ abs(a

_{i})

(5) If ∑ abs(a

_{i}) is convergent, then by Lemma 1 above, ∑ u

_{i}and ∑ v

_{i}are convergent.

We can use Lemma 1 since:

(a) ∑ abs(a

_{i}) is convergent and all abs(a

_{i}) ≥ 0.

(b) all u

_{i}and v

_{i}≥ 0 (see #3a)

(c) N=1, K=1 since all u

_{i}≤ abs(a

_{i}) [#4a] and all v

_{i}≤ abs(a

_{i}) [#4b]

(6) Using Lemma 2 above, we can conclude that ∑ (u

_{i}- v

_{i}) is also convergent.

(7) Then, since a

_{i}= u

_{i}- v

_{i}(see #3c), it follows that ∑ a

_{i}is also convergent.

QED

Theorem 4: if ∑ a

_{i}is conditionally convergent, then for any value L, it is possible to reorder a

_{i}and create an infinite sum ∑ a

_{o(i)}such that ∑ a

_{o(i)}= L and o(i) is an ordering function on i.

Proof:

(1) Let b

_{i}be defined such that:

if a

_{i}≥ 0, then b

_{i}= a

_{i}

if a

_{i}≤ 0, then b

_{i}= 0

(2) Let c

_{i}be defined such that:

if a

_{i}≥ 0, then c

_{i}= 0

if a

_{i}≤ 0, then c

_{i}= a

_{i}

(3) It follows that:

(a) a

_{i}= b

_{i}+ c

_{i}

(b) abs(a

_{i}) = b

_{i}- c

_{i}

(4) Let:

A

_{n}= ∑ (i=1,n) a

_{i}

B

_{n}= ∑ (i=1,n) b

_{i}

C

_{n}= ∑ (i=1,n) c

_{i}

(5) Let A

_{n}* = ∑ (i=1,n) abs(a

_{i})

(6) Then B

_{n}= (1/2)(A

_{n}+ A

_{n}*) since:

(a) If we add the two equations a

_{i}= b

_{i}+ c

_{i}(#3a) and abs(a

_{i}) = b

_{i}- c

_{i}(#3b), we get:

2*b

_{i}= a

_{i}+ abs(a

_{i})

(b) Since this is true for each term, we get: 2*B

_{n}= A

_{n}+ A

_{n}*

(c) This gives us:

B

_{n}= (1/2)(A

_{n}+ A

_{n}*)

(7) C

_{n}= (1/2)(A

_{n}- A

_{n}*) since:

(a) If we subtract abs(a

_{i}) = b

_{i}- c

_{i}(#3b) from a

_{i}= b

_{i}+ c

_{i}(#3a), we get:

2*c

_{i}= a

_{i}- abs(a

_{i})

(b) Since this is true for each term, we get: 2*C

_{n}= A

_{n}- A

_{n}*

(c) This gives us:

C

_{n}= (1/2)(A

_{n}- A

_{n}*)

(8) Since ∑ a

_{i}is conditionally convergent (see Definition 3 above), we know that:

∑ a

_{i}is convergent but ∑ abs(a

_{i}) is divergent.

(9) This means that ∑ b

_{i}is a divergent series of nonnegative terms. [See step #6]

(10) This also means that ∑ c

_{i}is a divergent series of nonpositive terms. [See step #7]

(11) Let n

_{1}be the least integer such that ∑ (i=1,n

_{1}) b

_{i}is greater than L

(12) Let n

_{2}be the least integer such that ∑ (i=1,n

_{1})b

_{i}+ ∑ (j=1,n

_{2}) c

_{j}is less than L.

(13) Let n

_{3}be the least integer such that ∑ (i=1,n

_{1}+ n

_{3})b

_{i}+ ∑ (j=1,n

_{2})c

_{j}is greater than L.

(14) We can likewise define all n

_{i}in a similar manner.

(15) We can now define a new series ∑ u

_{i}such that:

For i ≤ n

_{1}, let u

_{i}= b

_{i}

For j ≤ n

_{2}, let u

_{(n1+j)}= c

_{j}

For k ≤ n

_{3}, let u

_{(n1+n2+k)}= b

_{k}

And so on for the rest of the series.

(16) Let U

_{i}= ∑ u

_{i}

(17) We can see that:

U

_{n1}is greater than L

U

_{n1 + n2}is less than L

U

_{n1 + n2 + n3}is greater than L

(18) abs(U

_{n1}- L) is less than abs(u

_{n1}) since:

(a) U

_{n1}is greater than L which is greater than U

_{(n1-1)}

(b) abs(L - U

_{n1}) is less than abs(U

_{(n1-1)}- U

_{n1}) = abs(u

_{n1})

(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 n

_{1}and (n

_{1}+ n

_{2}), then:

U

_{n}- L is between U

_{(n1 + n2)}- L and U

_{n1}- L since:

(a) Assume n

_{1}less than n less than n

_{1}+ n

_{2}

(b) Since c

_{i}consists only of nonpositive numbers, we know that:

u

_{n1}≥ u

_{n}greater than u

_{(n1 + n2)}[Since u

_{(n1 + n2)}is negative and is the least number where ∑ u

_{i}is less than L.]

(c) This gives us that:

(U

_{n1}- L) ≥ (U

_{n}- L) which is greater than (U

_{(n1+n2)}- L)

(21) We can now conclude that abs(U

_{n-L}) is less than abs(u

_{n1}) + abs(u

_{(n1 + n2)}) since:

(a) abs(U

_{n}-L) ≤ abs(U

_{n1}- L) [#20c]

(b) abs(U(

_{n1+>n2}) - L) ≥ 0 [#12]

(c) So, abs(U

_{n}-L) ≤ abs(U

_{n1}- L) + abs(U(

_{n1+>n2}) - L)

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

abs(U

_{n}-L) is less than abs(u

_{n1}) + abs(u

_{(n1 + n2)})

(22) Since ∑ a

_{i}is convergent, there exists N such that:

if i is greater than N, abs(a

_{i}) is less than (1/2)ε

(23) There exists an integer m such that:

n

_{1}+ n

_{2}+ ... + n

_{m}≤ n is less than n

_{1}+ n

_{2m + nm+1}

This follows since n

_{i}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(U

_{n}- L) is less than abs(u

_{(n1 + n2 + ... + nm)}) + abs((u

_{(n1 + n2 + ... + nm + nm+1)})

(25) Further, we note that all abs(a

_{i}) is less than (1/2ε), we can conclude that:

abs(U

_{n}- L) is less than (1/2ε) + (1/2ε) = ε

(26) Hence, we have proved that L is also the limit for ∑ u

_{i}[See Definition 1 of Convergence above]

(27) In our building of ∑ u

_{i}, there is a strong possibility of zeros being added. (See step #15)

(28) Let v

_{i}be defined such that it has the same order as u

_{i}but only includes values where u

_{i}≠ 0.

(29) It is clear that v

_{i}is an infinite sum and ∑ v

_{i}= ∑ u

_{i}= L.

QED

Theorem 5: Rearrangeable Terms

if:

all a

_{i}≥ 0 and b

_{i}is a rearrangement of a

_{i}such that for all a

_{i}= b

_{o(i)}where o(i) is an ordering function on i.

then:

if ∑ a

_{i}converges, then ∑ b

_{i}does too.

if ∑ a

_{i}diverges, then ∑ b

_{i}also diverges.

Proof:

(1) Assume that:

b

_{1}= a

_{m1}

b

_{2}= a

_{m2}

b

_{3}= a

_{m3}

and so on

(2) Let B

_{n}= ∑ (i=1,n) b

_{i}and A

_{n}= ∑ (i=1,n) a

_{i}

(3) For a set m

_{1}, m

_{2}, ... , m

_{n}, let p = the largest of the integers.

(4) So, we have:

B

_{n}= ∑ (i=1,n) a

_{mi}≤ ∑ (i=1,p) a

_{i}= A

_{p}

(5) For A

_{n}, let k be the point where B

_{k}≥ A

_{n}.

(6) Then, we have:

A

_{n}= ∑ (i=1,n) a

_{i}≤ ∑ (i=1,k) b

_{i}= B

_{k}

(7) Assume that ∑ a

_{i}converges to L

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

_{n}≤ L

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

_{n}≥ L

(10) This shows that if ∑ a

_{i}converges to L, then ∑ b

_{i}also converges to L.

(11) If ∑ a

_{i}is divergent, then step #6 tells that ∑ b

_{i}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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