Combination Theorem for Sequences/Complex/Sum Rule/Proof 2
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Theorem
Let $\sequence {z_n}$ and $\sequence {w_n}$ be sequences in $\C$.
Let $\sequence {z_n}$ and $\sequence {w_n}$ be convergent to the following limits:
- $\ds \lim_{n \mathop \to \infty} z_n = c$
- $\ds \lim_{n \mathop \to \infty} w_n = d$
Then:
- $\ds \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$
Proof
Let $\epsilon > 0$ be given.
Then $\dfrac \epsilon 2 > 0$.
We are given that:
- $\ds \lim_{n \mathop \to \infty} z_n = c$
- $\ds \lim_{n \mathop \to \infty} w_n = d$
Let:
- $z_n = x_n + i y_n$
- $w_n = r_n + i s_n$
- $c = a + i b$
- $d = l + i m$
where each of $x_n, y_n, r_n, s_n, a, b, l, m \in \R$ are real.
By definition:
$\sequence {z_n}$ converges to the limit $c = a + i b$ if and only if both:
- $\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {x_n - a} < \epsilon \text { and } \size {y_n - b} < \epsilon$
$\sequence {w_n}$ converges to the limit $d = l + i m$ if and only if both:
- $\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {r_n - l} < \epsilon \text { and } \size {s_n - m} < \epsilon$
where $\size x$ denotes the absolute value of $x \in \R$.
Then:
\(\ds \lim_{n \mathop \to \infty} \paren {z_n + w_n}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {x_n + i y_n + r_n + i s_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {x_n + r_n + i \paren {y_n + s_n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {x_n + r_n} + i \lim_{n \mathop \to \infty} \paren {y_n + s_n}\) | Definition 1 of Convergent Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a + l} + i \paren {b + m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a + i b} + \paren {l + i m}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds c + d\) |
$\blacksquare$