The Algebraic Limit Theorem (ii)

[The Algebraic Limit Theorem] Let $\lim a_n = a$ and $\lim b_n = b$. Then,

$\lim (ca_n) = ca$ for all $c \in \mathbb{R}$;
$\lim (a_n + b_n) = a + b$;
$\lim (a_nb_n) = ab$;
$\lim (a_n/b_n) = a$;

Previous Definitions:

Problem Discussion

We’ll follow a similar approach to proving (i) in here For (ii) we want to show that $\lim (a_n + b_n) = a + b$. So we want to find $N$ such that for any $\epsilon > 0$,

$$ \begin{align*} \lvert a_n + b_n - (a + b) \rvert < \epsilon. \end{align*} $$

We can use the triangle inequality to bound the term:

$$ \begin{align*} \lvert a_n + b_n - (a + b) \rvert &= \lvert (a_n - a) + (b_n - b) \rvert \\ &\leq \lvert a_n - a \rvert + \lvert b_n - b \rvert. \end{align*} $$

So now the goal is to get the right hand side to be $\epsilon$. To do so we’re given that $\lim a_n = a$. This means that we can make $\lvert a_n - a \rvert$ be as small as we want and similarly we can make $\lvert b_n - b \rvert$ be as small as we want. So if we make each of them be $\epsilon/2$ then the terms will total to $\epsilon$ which is what we want. Formally, given that $\lim a_n = a$, then we know that for some $n \geq N_1$,

$$ \begin{align*} \lvert a_n - a \rvert < \frac{\epsilon}{2}. \end{align*} $$

Similarly, given that $\lim b_n = b$, then we know that for some $n \geq N_2$,

$$ \begin{align*} \lvert b_n - b \rvert < \frac{\epsilon}{2}. \end{align*} $$

You can see now if we replace these terms in the equation above we will get

$$ \begin{align*} \lvert (a_n - a) + (b_n - b) \rvert &\leq \lvert a_n - a \rvert + \lvert b_n - b \rvert = \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon. \end{align*} $$

$N$ can safely be set to the maximum of $N_1$ and $N_2$ to guarantee the above bounds.

Formal Proof

Let $\epsilon > 0$ be arbitrary. Since we want to bound the sum $\lvert (a_n - a) + (b_n - b) \rvert$, we apply the definition of the limit to each sequence using $\epsilon/2$. Since $\lim a_n = a$, then this means that for some $n \geq N_1$, we must have

$$ \begin{align*} \lvert a_n - a \rvert < \frac{\epsilon}{2}. \end{align*} $$

Similarly, since $\lim b_n = b$, then this means that for some $n \geq N_2$, we must have

$$ \begin{align*} \lvert b_n - b \rvert < \frac{\epsilon}{2}. \end{align*} $$

Therefore, choose a natural number $N$ satisfying $\max(N_1,N_2)$. We now verify that this choice has the desired property. Let $n \geq N$. Then,

$$ \begin{align*} \lvert (a_n - a) + (b_n - b) \rvert &\leq \lvert a_n - a \rvert + \lvert b_n - b \rvert \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align*} $$

as we wanted to show. $\blacksquare$

References

Understanding Analysis by Stephen Abbott