6  Sequences in \(\mathbb{C}\)

Definition 1: Convergent sequence in \(\mathbb{C}\)

We say that a sequence \(\left(a_{n}\right)\) in \(\mathbb{C}\) converges to \(a \in \mathbb{C}\), or equivalently has limit \(a\), denoted by \[ \lim_{n \rightarrow \infty} a_{n}=a \quad \mbox{ or } \quad a_n \to a \,, \] if it holds: \[ \forall \, \varepsilon>0 , \, \exists \, N \in \mathbb{N} \, \mbox{ s.t. } \, \forall \, n \geq N \,, \,\left|a_{n}-a\right| < \varepsilon \,. \]

If there exists \(a \in \mathbb{C}\) such that \(\lim _{n \rightarrow \infty} a_{n}=a\), we say that the sequence \(\left(a_{n}\right)\) is convergent.

Example 2

Question. Using Definition 2, prove that \[ \lim_{n \rightarrow \infty} \frac{(3+i) n-7 i}{n}=3+i \,. \]

Solution.

Part 1. Rough Work. Let \(\varepsilon >0\). We need to clarify for which values of \(n\) the following holds: \[ \left|\frac{(3+i) n-7 i}{n}-(3+i)\right| < \varepsilon \,. \] We have \[ \left|\frac{(3+i) n-7 i}{n}-(3+i)\right| = \frac{|-7 i|}{n} = \frac{7}{n} \,.\\ \] Therefore \[ \frac{7}{n} < \varepsilon \quad \iff \quad n > \frac{7}{\varepsilon} \,. \]

Part 2. Formal Proof. We want to prove that for all \(\varepsilon>0\) there exists \(N \in \mathbb{N}\) such that \[ \left|\frac{(3+i) n-7 i}{n}-(3+i)\right| < \varepsilon \,, \qquad \forall \, n \geq N \,. \] Let \(\varepsilon>0\). Choose \(N \in \mathbb{N}\) such that \[ N > \frac{7}{\varepsilon} \,. \] The above is equivalent to \[ \frac{7}{N} < \varepsilon \,. \] For \(n \geq N\) we have \[ \left|\frac{(3+i) n-7 i}{n}-(3+i)\right| = \frac{7}{n} \leq \frac{7}{N} < \varepsilon \,. \]

Definition 3: Bounded sequence in \(\mathbb{C}\)

A sequence \(\left(a_{n}\right)\) in \(\mathbb{C}\) is called bounded if there exists a constant \(M \in \mathbb{R}\), with \(M>0\), such that \[ \left|a_{n}\right| \leq M \,, \quad \forall \, n \in \mathbb{N} \,. \]

Theorem 4

If a sequence \(\left(a_{n}\right)\) in \(\mathbb{C}\) converges, then the sequence is bounded.

Definition 5: Divergent sequences in \(\mathbb{C}\)

We say that a sequence \(\left(a_{n}\right)\) in \(\mathbb{C}\) is divergent if it is not convergent.

Corollary 6

Let \(\left(a_{n}\right)\) be a complex sequence. If \(\left(a_{n}\right)\) is not bounded, then it is divergent.

6.1 Algebra of limits in \(\mathbb{C}\)

Theorem 7: Algebra of limits in \(\mathbb{C}\)

Let \(\left(a_{n}\right)\) and \(\left(b_{n}\right)\) be sequences in \(\mathbb{C}\). Suppose that \[ \lim_{n \rightarrow \infty} a_{n}=a \,, \quad \lim_{n \rightarrow \infty} b_{n}=b \,, \] for some \(a,b \in \mathbb{C}\). Then,

1. Limit of sum is the sum of limits: \[ \lim_{n \rightarrow \infty}\left(a_{n} \pm b_{n}\right)=a \pm b \]

2. Limit of product is the product of limits: \[ \lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right) = a b \]

3. If \(b_{n} \neq 0\) for all \(n \in \mathbb{N}\) and \(b \neq 0\), then \[ \lim_{n \rightarrow \infty} \left(\frac{a_{n}}{b_{n}}\right)=\frac{a}{b} \]

Example 8

Question. Compute the limit of \[ a_{n} = \frac{(2-i) n^{2}+6 i n-5-3 i}{(6+3 i) n^{2}+11 i} \,. \]

Solution. Factor \(n^2\), the largest power of \(n\) in the denominator, \[ a_{n}= \frac{(2-i)+\dfrac{6 i}{n}- \dfrac{5}{n^2}- \dfrac{3 i}{ n^{2}} }{(6+3 i)+\dfrac{11 i}{n^{2}}} \longrightarrow \frac{2-i}{6+3 i} \,, \] where we used the Algebra of Limits. Finally, \[ \frac{2-i}{6+3 i} =\frac{(2-i)(6-3 i)}{(6+3 i)(6-3 i)} =\frac{1}{5}-\frac{4}{15} i \,. \]

6.2 Convergence to zero

Theorem 9

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{C}\) and suppose that \[ \lim_{n \rightarrow \infty}\left|a_{n}\right|=0 \,. \] Then \[ \lim_{n \rightarrow \infty} a_{n}=0 \,. \]

Example 10

Question. Prove that \(a_n \to 0\), where \[ a_{n}=\left(\frac{1}{2}+\frac{1}{3} i\right)^{n} \,. \]

Solution. We have \[\begin{align*} \left|a_{n}\right| & = \left|\left(\frac{1}{2}+\frac{1}{3} i\right)^{n}\right| \\ & = \left|\frac{1}{2}+\frac{1}{3} i\right|^{n} \\ & = \left(\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{3}\right)^{2}}\right)^{n} \\ & = \left(\sqrt{\frac{13}{36}}\right)^{n} \,. \end{align*}\] Since \[ \left|\sqrt{\frac{13}{36}} \right| < 1 \,, \] by the Geometric Sequence Test for real sequences, we conclude that \[ \left|a_{n}\right| \rightarrow 0 \,. \] Hence \(a_n \to 0\) by Theorem 9.

Example 11

Question. Consider the sequence \[ a_{n}:=\frac{2 i \cos (3 n) n+(7-i) n^{2}}{3 n^{2}+2 i n+\sin (2 n)} \,. \] Prove that \[ \lim_{n \to \infty} a_n = \frac{7}{3}-\frac{1}{3} i \,. \]

Solution. We divide by the largest power in the denominator, to get \[ a_{n} = \frac{\dfrac{2 i \cos (3 n)}{ n} + (7-i) }{3+ \dfrac{2 i}{n} + \dfrac{\sin (2 n)}{n^{2}}} \,. \] Notice that \[ - 1 \leq \cos (3 n) \leq 1 \,, \quad \forall \, n \in \mathbb{N} \,, \] and thus \[ - \frac{2}{n} \leq \frac{2 \cos (3 n)}{n} \leq \frac{2}{n} \,, \quad \forall \, n \in \mathbb{N} \,. \] Since \[ -\frac{2}{n} \longrightarrow 0 \,, \quad \frac{2}{n} \longrightarrow 0 \,, \] by the Squeeze Theorem we conclude that also \[ \frac{2 \cos (3 n)}{n} \to 0 \,. \] In particular we have shown that \[ \left| \dfrac{2 i \cos (3 n)}{ n} \right| = \left| \dfrac{2 \cos (3 n)}{ n} \right| \to 0 \,. \] Using Theorem 9 we infer \[ \dfrac{2 i \cos (3 n)}{ n} \to 0 \,. \] Similarly, \[ - \frac{1}{n^{2}} \leq \frac{\sin (2 n)}{n^{2}} \leq - \frac{1}{n^{2}} \,, \quad \forall \, n \in \mathbb{N} \,. \] Since \[ - \frac{1}{n^{2}} \longrightarrow 0 \,, \quad \frac{1}{n^{2}} \longrightarrow 0 \,, \] by the Squeeze Theorem we conclude \[ \frac{\sin (2 n)}{n^2} \longrightarrow 0 \,. \] Finally, we have \[ \left| \frac{2i}{n} \right| = \frac{2}{n} \longrightarrow 0 \,, \] and therefore \[ \frac{2i}{n} \longrightarrow 0 \] by Theorem 9. Using the Algebra of Limits in \(\mathbb{C}\) we conclude \[ a_{n} = \frac{\dfrac{2 i \cos (3 n)}{ n} + (7-i) }{3+ \dfrac{2 i}{n} + \dfrac{\sin (2 n)}{n^{2}}} \longrightarrow \frac{ 0+ (7-i) }{3+0+0}=\frac{7}{3}-\frac{1}{3} i \,. \]

6.3 Geometric sequence Test and Ratio Test in \(\mathbb{C}\)

Theorem 12: Geometric sequence Test in \(\mathbb{C}\)

Let \(x \in \mathbb{C}\) and let \(\left(a_{n}\right)_{n \in \mathbb{N}}\) be the geometric sequence in \(\mathbb{C}\) defined by \[ a_{n}:=x^{n} \,. \] We have:

1. If \(|x|<1\), then \[ \lim_{n \to \infty} a_{n} = 0 \,. \]

2. If \(|x|>1\), then sequence \(\left(a_{n}\right)\) is unbounded, and hence divergent.

Example 13

Question. Prove that \(a_n \to 0\), where \[ a_{n}=\frac{(-1+4 i)^{n}}{(7+3 i)^{n}} \,. \]

Solution. We first rewrite \[ a_{n}=\frac{(-1+4 i)^{n}}{(7+3 i)^{n}}=\left(\frac{-1+4 i}{7+3 i}\right)^{n} \] Then, we compute \[\begin{align*} \left|\frac{-1+4 i}{7+3 i}\right| & =\frac{|-1+4 i|}{|7+3 i|} \\ & = \frac{\sqrt{(-1)^{2}+4^{2}}}{\sqrt{7^{2}+3^{2}}} \\ & = \frac{\sqrt{17}}{\sqrt{58}} \\ & = \sqrt{\frac{17}{58}} \\ & < 1 \, \end{align*}\] By the Geometric Sequence Test \(a_{n} \rightarrow 0\).

Example 14

Question. Prove that \(a_n\) diverges, where \[ a_{n}=\frac{(-5+12 i)^{n}}{(3-4 i)^{n}} \,. \]

Solution. We first rewrite \[ a_{n}=\frac{(-5+12 i)^{n}}{(3-4 i)^{n}}=\left(\frac{-5+12 i}{3-4 i}\right)^{n} \,. \] We compute \[\begin{align*} \left|\frac{-5+12 i}{3-4 i}\right| & = \frac{|-5+12 i|}{|3-4 i|} \\ & = \frac{\sqrt{5^{2}+(-12)^{2}}}{\sqrt{3^{2}+(-4)^{2}}} \\ & = \frac{13}{5} \\ & > 1 \,. \end{align*}\] By the Geometric Sequence Test, the sequence \(a_n\) diverges.

Example 15

Question. Prove that \(a_n\) diverges, where \[ a_{n}= \exp \left( \frac{i \pi}{2} n \right)\,. \]

Solution. We have \[ \left|a_{n}\right|=\left|e^{\frac{i \pi}{2} n}\right|=1 \,, \] and hence the Geometric Sequence Test cannot be applied. However, we can see that \[ a_n = (i,-1,-i, 1, i,-1,-i, 1, \ldots ) \,, \] that is, \(a_n\) assumes only the values \(\{i,-1,-i,1\}\), and each of them is assumed infinitely many times. Therefore \(a_n\) is oscillating, and thus divergent.

Theorem 16: Ratio Test in \(\mathbb{C}\)

Let \(\left(a_{n}\right)\) be a sequence in \(\mathbb{C}\) such that \[ a_{n} \neq 0 \,, \quad \forall \, n \in \mathbb{N} \,. \]

1. Suppose that the following limit exists: \[ L:=\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \,. \] Then,

   • If \(L<1\) we have \[ \lim_{n \to\infty} a_{n}=0 \,. \]

   • If \(L>1\), the sequence \(\left(a_{n}\right)\) is unbounded, and hence does not converge.

2. Suppose that there exists \(N \in N\) and \(L>1\) such that \[ \left|\frac{a_{n+1}}{a_{n}}\right| \geq L \,, \quad \forall \, n \geq N \,. \] Then the sequence \(a_n\) is unbounded, and hence does not converge.

Example 17

Question. Study the convergence / divergence of the sequence \[ a_{n}=\frac{(4-3 i)^{n}}{(2 n) !} \,. \]

Solution. We compute \[\begin{align*} \left|\frac{a_{n+1}}{a_{n}}\right| & =\left|\frac{(4-3 i)^{n+1}}{(2(n+1)) !} \frac{(2 n) !}{(4-3 i)^{n}}\right| \\ & = \frac{|4-3 i|^{n+1}}{|4-3 i|^{n}} \cdot \frac{(2 n) !}{(2 n+2) !} \\ & = \frac{|4 - 3i|}{(2 n+2)(2 n+1)} \\ & =\frac{\sqrt{4^{2}+(-3)^{2}}}{(2 n+2)(2 n+1)} \\ & = \frac{5}{(2 n+2)(2 n+1)} \\ & = \frac{ \dfrac{5}{n^{2}}}{ \left(2+ \dfrac{2}{n} \right) \left(2+ \dfrac{1}{n} \right) } \longrightarrow L = 0 \,. \end{align*}\] Since \(L=0 < 1\), by the Ratio Test in \(\mathbb{C}\) we infer \(a_{n} \to 0\).

6.4 Convergence of real and imaginary part

Theorem 18

Let \(\left(z_{n}\right)_{n \in \mathbb{N}}\) be a sequence in \(\mathbb{C}\). For \(n \in \mathbb{N}\), let \(a_{n}, b_{n} \in \mathbb{R}\) such that \[ z_{n}=a_{n}+b_{n} i \,. \] Let \(z=a+b i\), with \(a, b \in \mathbb{R}\). Then \[ \lim_{n \rightarrow \infty} z_{n}=z \quad \iff \quad \lim_{n \rightarrow \infty} a_{n}=a \,, \quad \lim_{n \rightarrow \infty} b_{n}=b \text {. } \]

Example 19

Question. Consider the complex sequence \[ z_{n}:=\frac{\left(4 n+3 n^{2} i\right)\left(2 n^{2}+i\right)}{5 n^{4}} \,. \] Show that \[ \lim_{n \to \infty} z_n = \frac65 i \,. \]

Solution. We find the real and imaginary parts of \(z_n\) \[\begin{align*} z_n & = \frac{\left(4 n+3 n^{2} i\right)\left(2 n^{2}+i\right)}{5 n^{4}} \\ & = \frac{8 n^{3}+4 n i + 6 n^{4} i + 3 n^{2}i^2}{5 n^{4}} \\ & = \frac{8 n^{3}-3 n^{2}}{5 n^{4}} + \frac{6 n^{4}+4 n}{5 n^{4}} i \\ & = a_n + b_n i \,. \end{align*}\] Using the Algebra of Limits for real sequences we have that \[\begin{align*} & a_n = \frac{8 n^{3}-3 n^{2}}{5 n^{4}}=\frac{ \dfrac{8}{n} - \dfrac{3}{n^2} }{5} \longrightarrow \frac{0-0}{5}=0 \,, \\ & b_n = \frac{6 n^{4}+4 n}{5 n^{4}}=\frac{6+ \dfrac{4}{n^3} }{5} \longrightarrow \frac{6+0}{5}=\frac{6}{5} \,. \end{align*}\] By Theorem 18 we conclude \[ \lim _{n \rightarrow \infty} z_{n} = \lim_{n \to \infty} a_n + i \, \lim_{n \to \infty} b_n = 0+\frac{6}{5} i=\frac{6}{5} i \,. \]