# Limit Sum/Difference Laws for Convergent Sequences

We will now proceed to specifically look at the limit sum and difference laws (law 1 and law 2 from the Limit of a Sequence page) and prove their validity.

## Sum Law for Convergent Sequences

Recall that the following law regarding two convergent sequences $\{ a_n \}$ and $\{ b_n \}$:

Law 1 (Addition Law of Convergent Sequences): If the limits of the sequences $\{ a_n \}$ and $\{ b_n \}$ are convergent, that is $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ then $\lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n = A + B$. |

The diagram above illustrates the idea of the sum law for convergent sequences. For example, consider the sequences $\left \{ \frac{1}{n} \right \}_{n=1}^{\infty}$ and $\left \{ \frac{1}{n^2} \right \}_{n=1}^{\infty}$. Both of these sequences are convergent and $\lim_{n \to \infty} \frac{1}{n} = 0$ and $\lim_{n \to \infty} \frac{1}{n^2} = 0$. Theorem 1 above says that $\lim_{n \to \infty} \left ( \frac{1}{n} + \frac{1}{n^2} \right ) = 0 + 0 = 0$. We will now formally prove this property to show that our example is true.

**Proof of Law 1:**Let $\{ a_n \}$ and $\{ b_n \}$ be convergent sequences. By the definition of a sequence being convergent, we know that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ for some $A, B \in \mathbb{R}$.

- Now let $\epsilon > 0$ be given, and recall that $\lim_{n \to \infty} a_n = A$ implies that $\forall \epsilon \: \exists N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \frac{\epsilon}{2}$. Similarly, $\lim_{n \to \infty} b_n = B$ implies that $\forall \epsilon \: \exists N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid b_n - B \mid < \frac{\epsilon}{2}$.

- We will choose $N = \mathrm{max} \{ N_1, N_2 \}$. By choosing the larger of $N_1$ and $N_2$, we ensure that the if $n ≥ \mathrm{max} \{ N_1, N_2 \}$, then $\mid a_n - A \mid < \frac{\epsilon}{2}$ and $\mid b_n - B \mid < \frac{\epsilon}{2}$. We now want to show that if $n ≥ N$, then $\mid (a_n + b_n) - (A + B) \mid < \epsilon$. By the triangle inequality we obtain that:

- Therefore $\lim_{n \to \infty} (a_n + b_n) = A + B$. $\blacksquare$

## Difference Law for Convergent Sequences

Law 2 (Difference Law of Convergent Sequences): If the limits of the sequences $\{ a_n \}$ and $\{ b_n \}$ are convergent, that is $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ then $\lim_{n \to \infty} (a_n - b_n) = \lim_{n \to \infty} a_n - \lim_{n \to \infty} b_n = A - B$. |

The proof of law 2 is very similar to the proof of law 1.

**Proof of Law 2:**Let $\{ a_n \}$ and $\{ b_n \}$ be convergent sequences. By the definition of a sequence being convergent, we know that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$ for some $A, B \in \mathbb{R}$.

- Now let $\epsilon > 0$ be given, and recall that $\lim_{n \to \infty} a_n = A$ implies that $\forall \epsilon \: \exists N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \frac{\epsilon}{2}$. Similarly, $\lim_{n \to \infty} b_n = B$ implies that $\forall \epsilon \: \exists N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid b_n - B \mid < \frac{\epsilon}{2}$.

- We will choose $N = \mathrm{max} \{ N_1, N_2 \}$. By choosing the larger of $N_1$ and $N_2$, we ensure that the if $n ≥ \mathrm{max} \{ N_1, N_2 \}$, then $\mid a_n - A \mid < \frac{\epsilon}{2}$ and $\mid b_n - B \mid < \frac{\epsilon}{2}$. We now want to show that if $n ≥ N$, then $\mid (a_n - b_n) - (A - B) \mid < \epsilon$. By the triangle inequality we obtain that:

- Therefore $\lim_{n \to \infty} (a_n - b_n) = A - B$. $\blacksquare$