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, ..., 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:

a₁ - b₁ = (a₁ - b₁)

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

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

(13) Let n₃ be the least integer such that ∑ (i=1,n₁ + n₃)b_i + ∑ (j=1,n₂)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 ≤ n₁, let u_i = b_i

For j ≤ n₂, let u_(n1+j) = c_j

For k ≤ n₃, 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 n₁ and (n₁ + n₂), then:

U_n - L is between U_{(n1 + n2)} - L and U_n1 - L since:

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

(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:

n₁ + n₂ + ... + n_m ≤ n is less than n₁ + 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:

b₁ = a_m1

b₂ = a_m2

b₃ = 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 m₁, m₂, ... , 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