Jeffrey Adachi


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Example Proof: Using the Definition of a Limit With Explicit Sequence and Limit

Using the Definition of Convergence

an converges to limit, L ⇔ ∀ 𝜀 > 0, ∃ N ∈ ℕ s.t. n ≥ N ⇒ |an - L| < 𝜀

Show that an = 1 + 1/n2 converges to the limit, L=1.

  1. The question is not "What is the limit of an?"
    The limit is given in the problem.

  2. The question is not "How does one calculate the limit of an?"
    From previous classes we probably remember that as n gets large, a/n2 approaches zero, so an approaches 1 + 0 = 1.

  3. This is typical of a real analysis class.
    We already know how to calculate the limit.
    This class is about how to prove that our answer is correct.

  4. Proving things we know to be true allows us to focus on the thinking process for proofs, rather than the result being proven.
    But it can seem pointless to spend so much effort proving what is known (and possibly obvious).

Thought Process:

The definition includes, "∀ 𝜀 > 0, ∃ N s.t. ...", so for any positive 𝜀, we need to show that an N exists with the described property (n ≥ N ⇒ |an - L| < 𝜀).
A direct way to do this is to write N as an appropriate function of 𝜀 that is defined for all 𝜀 > 0.

So our task is to find a function N(𝜀) s.t.
n ≥ N(𝜀) ⇒ |an - L| < 𝜀

As a first step, we might ask whether any value of n satisfied |an - L| < 𝜀.

|an - L| < 𝜀
| 1 + 1/n2 - 1 | < 𝜀 (by definition of an and L)
| 1 - 1 + 1/n2 | < 𝜀 (arithmetic: a + b + c = a + c + b)
| 0 + 1/n2| < 𝜀 (arithmetic: 1 - 1 = 0)
| 1/n2| < 𝜀 (arithmetic: 0 + a = a)
1/n2 < 𝜀 (by definition of absolute value, |x| = x for non-negative x)
n2 > 1/𝜀 (arithmetic: a > b > 0 ⇒ 1/a < 1/b )
n > sqrt(1/𝜀) (arithmetic: a > b > 0 ⇒ sqrt(a) > sqrt(b))

Every integer n > sqrt(1/𝜀) will satisfy |an - L| < 𝜀.

This result is almost in the form we were looking for.
n > sqrt(1/𝜀) |an - L| < 𝜀 what we have so far
n ≥ N ∈ ℕ |an - L| < 𝜀 what is in the defintition

It is tempting to set N = sqrt(1/𝜀) and think we were done. But there are two problems:
  1. The upper expression uses ">" and the lower expression uses "≥".
  2. N must be an integer for all 𝜀, and in general, sqrt(1/𝜀) is not an integer.

We can force integrality by using ceiling(sqrt(1/𝜀)) and change the ">" to "≥" by adding one.
So we can choose N = 1 + ceiling(sqrt(1/𝜀)). Then

n ≥ N ∈ ℕ n ≥ 1 + ceiling(sqrt(1/𝜀))
n > ceiling(sqrt(1/𝜀))
n > sqrt(1/𝜀)
|an - L| < 𝜀

So, for any 𝜀>0, if we choose N(𝜀) = 1 + ceiling(sqrt(1/𝜀)), then

n≥N ⇒ |an - L|< 𝜀

NOTE: All the preceeding work is not a proof, it is an example of the scratch work needed to write the proof.

Writing out the Proof:

Let an = 1 + 1/n2 and L = 1.
∀ 𝜀 > 0, ∃ N(𝜀) = 1 + ceiling(sqrt(1/𝜀)) ∈ ℕ s.t.

n ≥ N |an - L| = |1 + 1/n2 - 1| by definition of an and choice of L
= |1 - 1 + 1/n2| arithmetic: commutativity of addition, a + b = b + a
= |0 + 1/n2| arithmetic: additive inverse, a - a = 0
= |1/n2| arithmetic: additive identity, 0 + a = a
= 1/n2 arithmetic: since n>0 so is 1/n2, |a| = a if a is positive
1/N2 since n ≥ N > 0, and a ≥ b > 0 ⇒ 1/a ≤ 1/b
1/(1+sqrt(1/𝜀))2 since N > 1+sqrt(1/𝜀)
< 1/sqrt(1/𝜀)2 since 1+sqrt(1/𝜀) > sqrt(1/𝜀)), and a > b > 0 ⇒ 1/a < 1/b
= 1/(1/𝜀) arithmetic: sqrt(a2) = |a|, and |a| = a for positive a
= 𝜀 arithmetic: property of the multiplicative inverse
n ≥ N |an - L| < 𝜀

Therefor, an = 1 + 1/n2 converges to limit L=1, by defintion.

Key Takeaways:

  1. The order in which the proof is presented is not the order in which the proof is constructed.
    Construction: Start with the conditions from the definition, then find N(𝜀) at the end.
    Presentation: N(𝜀) is given at the beginning, and then we show the conditions of the defintion are satisfied.
    If one reads only the proof, key parts (like 1 + ceiling(sqrt(1/𝜀))) seem to appear out of thin air.
    This is part of what makes studying proofs in a book difficult.
    The way to study the proof is to try the proof on your own and figure out the thought process that motivates the key parts.

  2. The proof demonstrates that for any 𝜀>0, there is an integer N with the required properties.
    The existence of an appropriate N was established by writing N explicitly as a function of 𝜀.
    Other proofs might might establish that such an N exists without knowing the functional form.

  3. Multiple basic properties of the real numbers may be needed to justify even what look like trivial arithmetic manipulations.

  4. A note on notation:
    The inequalities above are equivalent to

    n ≥ N ⇒
    |an - L| = |1 + 1/n2 - 1| = |1 - 1 + 1/n2| = |0 + 1/n2|= |1/n2| = 1/n2 ≤ 1/N2 ≤ etc.

    But the multiple line version allows associating a reason with each step.
    Also, I find it easier to see what was done in each step.
    And shouldn't clear communication the goal of writing proofs?
    This organization makes it easier for the student to find mistakes and easier for the grader to give due credit.

  5. A note on level of detail.
    Depending on the enthusiasm of your grader, you may be allowed to skip some of the steps shown above.

    e.g. |1 + 1/n2 - 1| = 1/n2

    Writing out every step is safer.
    How many mistakes in Calculus were due to doing arithmetic (incorrectly) in your head?
    The more steps you combine, the more difficult it will be for you to review your work for mistakes and the more difficult it wil be for the grader to follow what you did.
    Each student will have to evaluate whether the risk is worth it.

    If you ever had trouble following a proof in a book, consider trying to make your proofs easy to follow.

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