Limit of a Sequence
The limit of a sequence as n→∞ is expressed as $$ \lim_{n \rightarrow \infty} a_n = l $$
If the limit exists, the sequence is said to be a regular sequence.
A regular sequence can be:
- divergent (or infinite) if its limit equals ±∞.
- convergent if it approaches a finite number.
- infinitesimal if it converges to zero.
Note. If a sequence doesn’t have a limit, it’s called a non-regular sequence.
A Practical Example
Example 1
This sequence is bounded and converges to 1.
$$ \lim_{n \rightarrow \infty} \frac{n-1}{n} = 1 $$
In this case, the limit is l = 1, and the general term is an = (n-1)/n.
By the definition of the limit of a convergent sequence, for any ε>0, there exists some value v such that:
$$ l - \epsilon < a_n < l + \epsilon \quad \forall n > v $$
This inequality can also be written in a shorter, equivalent form:
$$ -\epsilon < a_n - l < \epsilon $$
$$ | a_n - l | < \epsilon $$
Substituting l = 1 and an = (n-1)/n gives:
$$ \left| \frac{n-1}{n} - 1 \right| < \epsilon $$
Let’s simplify the expression inside the absolute value:
$$ \left| \frac{n-1 - n}{n} \right| < \epsilon $$
$$ \left| \frac{-1}{n} \right| < \epsilon $$
$$ \frac{1}{n} < \epsilon $$
Finally, solving for n:
$$ \frac{1}{\epsilon} < n $$
This confirms that for every ε>0, there’s a threshold v = 1/ε beyond which the inequality holds for all n.
Therefore, the sequence is bounded and converges to 1:
$$ \lim_{n \rightarrow \infty} \frac{n-1}{n} = 1 $$
Note. You can also see this clearly by looking at the first few terms of the sequence: $$ a_n = 0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots $$ or in decimal form: $$ a_n = 0, 0.5, 0.66, 0.75, 0.8, \dots $$
Example 2
Let’s test whether the following sequence converges to 1:
$$ \lim_{n \rightarrow \infty} \frac{1}{n} = 1 $$
Here, l = 1 and an = 1/n.
$$ | a_n - l | < \epsilon $$
$$ \left| \frac{1}{n} - 1 \right| < \epsilon $$
We can rewrite the absolute value as:
$$ 1 - \frac{1}{n} < \epsilon $$
Solving for n:
$$ - \frac{1}{n} < \epsilon - 1 $$
$$ \frac{1}{n} > 1 - \epsilon $$
This condition doesn’t hold for every ε>0, as required by the definition of the limit of a sequence.
Example. For ε = 1, the inequality is satisfied for any natural number n: $$ \frac{1}{n} > 0 $$ It also holds for ε > 1. For instance, with ε = 10: $$ \frac{1}{n} > -9 $$ However, if we choose ε < 1, it no longer holds. For example, with ε = 1/2: $$ \frac{1}{n} > \frac{1}{2} $$ which is only true when n = 1, and fails for n = 2 and any n > 1.
This shows that the sequence does not converge to 1:
$$ \lim_{n \rightarrow \infty} \frac{1}{n} \ne 1 $$
Uniqueness of the Limit
A convergent sequence can have only one limit.
A sequence cannot approach two or more different values at the same time.
Proof
Suppose, for the sake of contradiction, that a convergent sequence has two distinct limits l1 and l2 where l1 ≠ l2.
By definition, for any ε>0:
$$ \exists v_1 \:\: \text{such that } |a_n - l_1| < \epsilon \quad \forall n > v_1 $$
$$ \exists v_2 \:\: \text{such that } |a_n - l_2| < \epsilon \quad \forall n > v_2 $$
Let’s choose ε as follows:
$$ \epsilon = \frac{|l_1 - l_2|}{2} $$
Then:
$$ 2\epsilon = |l_1 - l_2| $$
$$ \epsilon + \epsilon = |l_1 - l_2| $$
Important. This last equality is crucial because it will be used at the end of the proof.
Let’s take the maximum of v1 and v2:
$$ v = \max(v_1, v_2) $$
Using this value ensures both conditions remain true:
$$ \exists v_1 \:\: \text{such that } |a_n - l_1| < \epsilon \quad \forall n > v $$
$$ \exists v_2 \:\: \text{such that } |a_n - l_2| < \epsilon \quad \forall n > v $$
Note. If n is greater than v, then it’s also greater than both v1 and v2: $$ n > v $$ $$ v \ge v_1, v_2 $$ For instance, if v2 > v1, then v = v2. So if n > v, it follows that n > v1 as well, since v1 < v.
Adding the two inequalities gives us:
$$ |a_n - l_1| + |a_n - l_2| < \epsilon + \epsilon $$
Important. This equation will also be essential for the conclusion of the proof.
Now, let’s examine the absolute difference between the two limits:
$$ | l_1 - l_2 | $$
We can insert and subtract an inside the absolute value:
This yields an equivalent expression:
$$ | l_1 - l_2 | = | l_1 - l_2 + a_n - a_n | $$
Grouping terms, we get:
$$ | l_1 - l_2 | = | ( l_1 - a_n ) + (a_n - l_2 ) | $$
By the triangle inequality, we have:
$$ | l_1 - l_2 | = | ( l_1 - a_n ) + (a_n - l_2 ) | \le | l_1 - a_n | + | a_n - l_2 | $$
Note. Recall the rule: $$ |a + b| \le |a| + |b| $$
We can reverse the terms inside each absolute value without changing the result:
$$ | l_1 - l_2 | = | ( l_1 - a_n ) + (a_n - l_2 ) | \le | a_n - l_1 | + | a_n - l_2 | $$
Note. Knowing that $$ |a_n - l_1| + |a_n - l_2| < \epsilon + \epsilon $$ and $$ \epsilon + \epsilon = |l_1 - l_2| $$
$$ | l_1 - l_2 | = | ( l_1 - a_n ) + (a_n - l_2 ) | < \epsilon + \epsilon $$
$$ | l_1 - l_2 | = | ( l_1 - a_n ) + (a_n - l_2 ) | < |l_1 - l_2| $$
$$ | l_1 - l_2 | < |l_1 - l_2| $$
But this is a contradiction.
This proves that a convergent sequence cannot have more than one limit.
And so on.
