Double Sequence Limits: UNIZOR.COM - Math4Teens - Calculus - Limit of Se...

Notes to a video lecture on http://www.unizor.com

Double Limits

Sometimes we are interested in sequences that depend on two natural numbers like {a_m,n}.
For example,
a_m,n=arctan(m)+2⁻ⁿ.

Now, if m→∞ and n→∞, how to calculate the limit of a_m,n?
It might be
lim_m→∞ [lim_n→∞ (a_m,n)]
or
lim_n→∞ [lim_m→∞ (a_m,n)]
depending on which index, m or n, we will use to go to a limit first, and there is no guarantee that the answers will be the same.

Let's check both methods.
lim_n→∞ (a_m,n) =
= lim_n→∞ (arctan(m) + 2⁻ⁿ) =
= arctan(m)
lim_m→∞ (arctan(m)) = π/2

If we change the order of limits,
lim_m→∞ (a_m,n) =
= lim_m→∞ (arctan(m) + 2⁻ⁿ) =
= π/2 + 2⁻ⁿ
lim_n→∞ (π/2 + 2⁻ⁿ) = π/2

As you see, the results are the same.

In some way, the equality of these two limits might be considered similar to a standard accounting procedure of checking the calculations.
Imagine an M⨯N matrix with numbers.
If you summarize the numbers within each of M rows with row #i having a sum R_i and summarize all these row totals by i from 1 to M, you will get a total of all numbers in an original table.
If you summarize the numbers within each of N columns with column #j having a sum S_j and summarize all these column totals by j from 1 to N, you should get exactly the same total of all numbers.
If these two calculated totals are not equal, that is if
Σ_i∈[1,M] R_i ≠ Σ_j∈[1,N] S_j
then your calculations are wrong.

The equality of two limits in the example above is not coincidental.
We shall prove the following theorem.

Theorem

IF
lim_n→∞ (a_m,n) = b_m
lim_m→∞ (b_m) = B
lim_m→∞ (a_m,n) = c_n
lim_n→∞ (c_n) = C
THEN
B = C

Proof

Assume, B≠C and |B−C|=d.

Since lim_m→∞(b_m)=B, there exists some natural number M₁ such that
|B−b_m| < d/4 for all m>M₁

Since lim_n→∞(a_m,n)=b_m, there exists some natural number N₁ such that
|a_m,n−b_m| < d/4 for all n>N₁

Then, for all pairs m,n such that m>M₁ and n>N₁ both following inequalities are true:
|B − b_m| < d/4 and
|b_m − a_m,n| < d/4
Therefore,
|B−a_m,n| < d/2
That is, our double sequence a_m,n with both indices m and n sufficiently large (m>M₁ and n>N₁) is closer to number B than d/2.

Absolutely similarly we can prove that this same sequence a_m,n with both indices sufficiently large (m>M₂ and n>N₂) is closer to number C than d/2.
Here is a proof in mathematical symbolics.
lim_n→∞ (c_n) = C ∴
∴ ∃N₂: |C−c_n| < d/4 ∀n>N₂
lim_m→∞ (a_m,n) = c_n ∴
∴ ∃M₂: |c_n−a_m,n| < d/4 ∀m>M₂
m>M₂ ∧ n>N₂ ∴
∴ |C−c_n| < d/4 ∧ |c_n−a_m,n| < d/4 ∴
∴ |C−a_m,n| < d/2

Choosing M=max(M₁,M₂) and N=max(N₁,N₂) we see that all a_m,n with m>M and n>N should be closer to B than d/2 and at the same time should be closer to C than d/2.
With the distance between B and C equal to d it's impossible.
We came to contradiction that signifies that our initial assumption that B≠C is wrong.
Therefore, B=C.

End of proof.