2.1 Sequences
Definition 2.1. A sequence \(\{z_n\}\) is a function where the domain is the set of positive integers and
whose range is a subset of the complex numbers.
If \(\lim \limits _{n \rightarrow \infty } z_n = L\), we say the sequence \(\{z_n\}\) is convergent. The number \(L\) is called the limit of \((z_n)\)
Definition 2.2. A sequence \(\{z_n\}\) converges to the number \(L\) if \(\forall \, \varepsilon > 0\) and \(N\) exist such that \(\,\left |z_n - L\right | < \varepsilon \,\) whenever \(n<N\).
A sequence that is not convergent is said to be divergent.
Example 2.3. The sequence \(\,\displaystyle {\Big \{\frac {i^{n + 1}}{n}\Big \}}\,\) converges, since \(\, \lim \limits _{n\rightarrow \infty } \frac {i^{1 + n}}{n} = 0\).
As we see from this \(\,\displaystyle {-1,\, \frac {-i}{2}, \, \frac {1}{3},\,\frac {i}{1}, \,, \frac {-1}{5}}\).
Theorem 2.4. A sequence \(\{z_n\}\) converges to a complex number \(L = a + ib\) if and only if \(\operatorname {Re}(z_n)\) converges to \(\operatorname {Re}(L)= a\) and \(\operatorname {Im}(z_n)\) converges to \(\operatorname {Im}(L) = b\).
Proof. Assume that \(z_n \longrightarrow L\). Then \(\forall \, \varepsilon > 0\, \exists \, N_{\varepsilon } \ni \left |z_n - L\right | < \varepsilon \, \forall \, n> N_{\varepsilon }\). But
\[\left |\operatorname {Re}(z_n - L)\right | \leq \left |z_n - L\right |< \varepsilon \]
That is, \(\operatorname {Re}(z_n) \longrightarrow \operatorname {Re}(L)\). Similarly, one can show that \(\operatorname {Im}(z_n) \longrightarrow \operatorname {Im} (L)\).
Conversely, assume that (abbreviating \(x = \operatorname {Re} (L)\) and \(y = \operatorname {Im} (L))\)
\(\, \forall \, \varepsilon > 0\, \exists \, N_{\varepsilon } \ni \left |x_n - x\right |,\,\left |y_n - y\right | < \frac {\varepsilon }{2}\,\forall \,n>N_{\varepsilon }\). Then by the triangle inequality
\[ \left |z_n - L\right | \leq \left |x_n - x\right | + \left |y_n - y\right | < \frac {\varepsilon }{2} + \frac {\varepsilon }{2} = \varepsilon \]
\(\implies \, z_n \longrightarrow L\) as \(n \longrightarrow \infty .\) □
Solution \begin {align*} \text {From}\qquad z_n & = \frac {3 + ni}{n + 2ni} = \frac {(3 + ni) (n-2ni)}{n^2 + 4n^2}\\\\ & = \frac {2n^2 + 3n}{5n^2} + i\,\frac {n^2 - 6n}{5n^2} \end {align*}
we see that
\(\operatorname {Re}(z_n) = \frac {2n^2 + 3n}{5n^2} = \frac {2}{5} + \frac {3}{5n} \longrightarrow \frac {2}{5}\) as \(n\longrightarrow \infty \)
\(\operatorname {Im}(z_n) = \frac {n^2 - 6n}{5n^2} = \frac {1}{5} - \frac {6}{5n}\longrightarrow \frac {1}{5}\) as \(n \longrightarrow \infty \)
Hence \(\, z_n \longrightarrow \frac {2}{5} + i\,\frac {1}{5}\)
Definition 2.6. A sequence \(\{z_n\}_{n\geq 1}\) is said to be Cauchy if \(\forall \, \varepsilon > 0, \exists \, n_0 \in \mathbb {N} \ni n,m \geq n_0,\,\left |z_m - z_n\right | < \varepsilon \).
Proof. Write \(z_n=x_n+iy_n\). The whole proof rests on the two inequalities \[\left |x_n-x_m\right |\leq \left |z_n-z_m\right |,\qquad \left |y_n-y_m\right |\leq \left |z_n-z_m\right |,\] \[\left |z_n-z_m\right |\leq \left |x_n-x_m\right |+\left |y_n-y_m\right |,\] which say that a complex sequence is Cauchy exactly when both of its real sequences are.
If \(z_n\rightarrow z=x+iy\) then for \(n,m\) large \(\left |z_n-z_m\right |\leq \left |z_n-z\right |+\left |z-z_m\right |\) is small, so \((z_n)\) is Cauchy. That direction needs nothing about \(\mathbb {C}\) in particular.
Conversely, if \((z_n)\) is Cauchy then by the first pair of inequalities \((x_n)\) and \((y_n)\) are Cauchy sequences of real numbers, hence convergent, by the completeness of \(\mathbb {R}\). Say \(x_n\rightarrow x\) and \(y_n\rightarrow y\). Then \[\left |z_n-(x+iy)\right |\leq \left |x_n-x\right |+\left |y_n-y\right | \longrightarrow 0 ,\] so \(z_n\rightarrow x+iy\).
The content is inherited: \(\mathbb {C}\) is complete because \(\mathbb {R}\) is, and the passage between them is made by those inequalities. □
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