By reading this material we can easily understand the topic.

Course Material 2.3 Algebra of Convergent Sequences

Theorem.

Suppose 𝑥𝑛 → 𝑥 𝑎𝑛𝑑 𝑦𝑛 → 𝑦. Then

1. 𝑥𝑛 + 𝑦𝑛 → 𝑥 + 𝑦
2. 𝛼𝑥𝑛 → 𝛼𝑥
3. 𝑥𝑛 ∙ 𝑦𝑛 → 𝑥 ∙ 𝑦
1 1
4. → 𝑥 provided 𝑥 ≠ 0
𝑥𝑛

Proof.

1. Given that 𝑥𝑛 → 𝑥 𝑎𝑛𝑑 𝑦𝑛 → 𝑦 then given 𝜖 > 0 there exists a positive integer 𝑁1 such
𝜖
that for all 𝑛 ≥ 𝑁1 |𝑥𝑛 − 𝑥 | < 2.

Also, as 𝑦𝑛 → 𝑦 , given 𝜖 > 0 there exists a positive integer 𝑁2 such that for all 𝑛 ≥ 𝑁2
𝜖
|𝑦𝑛 − 𝑦| < .
2

Then if 𝑁 = max{𝑁1 , 𝑁2 } for all 𝑛 ≥ 𝑁 we have 𝑛 ≥ 𝑁1 and 𝑛 ≥ 𝑁2 so that the above
two inequalities will hold
Let 𝑛 ≥ 𝑁 then,
|(𝑥𝑛 + 𝑦𝑛 ) − (𝑥 + 𝑦)| = |𝑥𝑛 − 𝑥 + 𝑦𝑛 − 𝑦|
≤ |𝑥𝑛 − 𝑥 | + |𝑦𝑛 − 𝑦|
𝜖 𝜖
< + =𝜖
2 2

This shows that 𝑥𝑛 + 𝑦𝑛 → 𝑥 + 𝑦
2. Proof is left
3. Since {𝑥𝑛 } converges, it is bounded. Then we can find 𝐶 such that |𝑥𝑛 | ≤ 𝐶 for all 𝑛. Also
𝐶 is choosen such that |𝑦| ≤ 𝐶
As in (1) given 𝜖 > 0 there exists a positive integer 𝑁 such that for all 𝑛 ≥ 𝑁, we have
𝜖 𝜖
|𝑥𝑛 − 𝑥 | < and |𝑦𝑛 − 𝑦| < 2𝐶
2𝐶

|𝑥𝑛 𝑦𝑛 − 𝑥𝑦| = |𝑥𝑛 𝑦𝑛 − 𝑥𝑛 𝑦 + 𝑥𝑛 𝑦 − 𝑥𝑦|
≤ |𝑥𝑛 𝑦𝑛 − 𝑥𝑛 𝑦| + |𝑥𝑛 𝑦 − 𝑥𝑦|
≤ |𝑥𝑛 ||𝑦𝑛 − 𝑦| + |𝑦||𝑥𝑛 − 𝑥 |
≤ 𝐶 |𝑦𝑛 − 𝑦| + 𝐶 |𝑥𝑛 − 𝑥 |