Combination Theorem for Sequences/Real

Theorem
Let $$X$$ be one of the standard number fields $$\Q, \R, \C$$.

Let $$\left \langle {x_n} \right \rangle$$ and $$\left \langle {y_n} \right \rangle$$ be sequences in $X$.

Let $$\left \langle {x_n} \right \rangle$$ and $$\left \langle {y_n} \right \rangle$$ be convergent to the following limits:


 * $$\lim_{n \to \infty} x_n = l, \lim_{n \to \infty} y_n = m$$

Let $$\lambda, \mu \in X$$.

Then the following results hold:

Sum Rule

 * $$\lim_{n \to \infty} \left({x_n + y_n}\right) = l + m$$

Multiple Rule

 * $$\lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l$$

Combined Sum Rule

 * $$\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$$

Product Rule

 * $$\lim_{n \to \infty} \left({x_n y_n}\right) = l m$$

Quotient Rule

 * $$\lim_{n \to \infty} \frac {x_n} {y_n} = \frac l m$$, provided that $$m \ne 0$$

Proof of Sum Rule
We need to show that $$\lim_{n \to \infty} \left({x_n + y_n}\right) = l + m$$.

Let $$\epsilon > 0$$ be given. Then $$\frac \epsilon 2 > 0$$.

Since $$\lim_{n \to \infty} x_n = l$$, we can find $$N_1$$ such that $$\forall n > N_1: \left|{x_n - l}\right| < \frac \epsilon 2$$.

Similarly, since $$\lim_{n \to \infty} y_n = m$$, we can find $$N_2$$ such that $$\forall n > N_2: \left|{y_n - m}\right| < \frac \epsilon 2$$.

Now let $$N = \max \left\{{N_1, N_2}\right\}$$.

Then if $$n > N$$, both the above inequalities will be true.

Thus $$\forall n > N$$:

$$ $$ $$

Hence $$\lim_{n \to \infty} \left({x_n + y_n}\right) = l + m$$.

Proof of Multiple Rule
Let $$\epsilon > 0$$.

We need to find $$N$$ such that $$\forall n > N: \left|{\lambda x_n - \lambda l}\right| < \epsilon$$.

If $$\lambda = 0$$ the result is trivial.

So, assume $$\lambda \ne 0$$.

Then $$\left|{\lambda}\right| > 0$$ from the definition of the modulus of $$\lambda$$.

Hence $$\frac {\epsilon} {\left|{\lambda}\right|} > 0$$.

We have that $$x_n \to l$$ as $$n \to \infty$$.

Thus it follows that $$\exists N: \forall n > N: \left|{x_n - l}\right| < \frac {\epsilon} {\left|{\lambda}\right|}$$.

That is, $$\forall n > N: \left|{\lambda}\right| \left|{x_n - l}\right| < \epsilon$$.

But we have:

$$ $$

Hence $$\lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l$$.

Proof of Combined Sum Rule
We need to show that $$\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$$.

From the Multiple Rule above, we have:
 * $$\lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l$$
 * $$\lim_{n \to \infty} \left({\mu y_n}\right) = \mu m$$

The result now follows directly from the Sum Rule above:
 * $$\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$$

Proof of Product Rule
We need to show that $$\lim_{n \to \infty} \left({x_n y_n}\right) = l m$$.

Since $$\left \langle {x_n} \right \rangle$$ converges, it is bounded by Convergent Sequence is Bounded.

Suppose $$\left|{x_n}\right| \le K$$ for $$n = 1, 2, 3, \ldots$$.

Then:

$$ $$ $$ $$ $$

But $$x_n \to l$$ as $$n \to \infty$$.

So $$\left|{x_n - l}\right| \to 0$$ as $$n \to \infty$$ from Convergent Sequence Minus Limit.

Similarly $$\left|{y_n - m}\right| \to 0$$ as $$n \to \infty$$.

From the Combined Sum Rule above, $$\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$$, $$z_n \to 0$$ as $$n \to \infty$$.

The result follows by the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences).

Proof of Quotient Rule
We need to show that $$\lim_{n \to \infty} \frac {x_n} {y_n} = \frac l m$$, provided that $$m \ne 0$$.

As $$y_n \to m$$ as $$n \to \infty$$, it follows from Modulus of Limit that $$\left|{y_n}\right| \to \left|{m}\right|$$ as $$n \to \infty$$.

As $$m \ne 0$$, it follows from the definition of the modulus of $$m$$ that $$\left|{m}\right| > 0$$.

From Sequence Converges to Within Half Limit, we have $$\exists N: \forall n > N: \left|{y_n}\right| > \frac {\left|{m}\right|} 2$$.

Now, for $$n > N$$, consider:

$$ $$

By the above, $$m x_n - y_n l \to ml - ml = 0$$ as $$n \to \infty$$.

The result follows by the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences).