# Combination Theorem for Sequences/Normed Division Ring/Sum Rule

## Theorem

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.

Let $\sequence {x_n}$, $\sequence {y_n}$ be sequences in $R$.

Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limits:

$\ds \lim_{n \mathop \to \infty} x_n = l$
$\ds \lim_{n \mathop \to \infty} y_n = m$

Then:

$\sequence {x_n + y_n}$ is convergent and $\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$

## Proof

Let $\epsilon > 0$ be given.

Then $\dfrac \epsilon 2 > 0$.

Since $\sequence {x_n}$ is convergent to $l$, we can find $N_1$ such that:

$\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon 2$

Similarly, for $\sequence {y_n}$ we can find $N_2$ such that:

$\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon 2$

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

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

Thus $\forall n > N$:

 $\ds \norm {\paren {x_n + y_n} - \paren {l + m} }$ $=$ $\ds \norm {\paren {x_n - l} + \paren {y_n - m} }$ $\ds$ $\le$ $\ds \norm {x_n - l} + \norm {y_n - m}$ Norm Axiom $\text N 3$: Triangle Inequality $\ds$ $<$ $\ds \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$

Hence:

$\sequence {x_n + y_n}$ is convergent to $l + m$.

$\blacksquare$