4.1 Complex Sequences

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We find limits of complex sequences.

A complex sequence $\{c_n\}$ is a function from the natural numbers to the complex numbers, $\C$ .

A complex sequence $\{c_n\}$ is said to converge to a complex number $c$ if given $\epsilon >0$ , there exists $N \in \N$ such that $|c_n - c| < \epsilon \quad \text {whenever} \quad n>N$ If no such complex number $c$ exists, then we say the sequence $\{c_n\}$ diverges.

example 1 Show that the sequence $\displaystyle \left \{\frac {i^n}{n}\right \}$ converges to $0$ .
Let $\epsilon >0$ and let $N > 1/\epsilon$ . Then for $n> N$ we have $\left |\frac {i^n}{n} -0\right | = \frac {1}{n} < \frac {1}{N} < \epsilon$ Hence, the sequence $\displaystyle \left \{\frac {i^n}{n}\right \}$ converges to $0$ .

(problem 1a) Show that the sequence $\displaystyle \left \{\frac {i^n}{2^n}\right \}$ converges to $0$ .

Let $\epsilon >0$

Let $N > -\log _2\epsilon$

If $n>N$ then $2^n > 2^{-\log _2\epsilon }$

(problem 1b) Show that the sequence $\displaystyle \left \{\frac {n+i}{in}\right \}$ converges to $-i$ .

Simplify $\displaystyle \left |\frac {n+i}{in} - (-i) \right |$

Here is a video solution to problem 1b:

The next proposition shows that convergence of complex sequences can be determined by the convergence of its real and imaginary parts.

A complex sequence $\{c_n\} = \{a_n + ib_n\}$ converges to $c = a+bi$ if and only if the real sequences $\{a_n\}$ and $\{b_n\}$ converge to $a$ and $b$ respectively.

Proof: $\left (\Rightarrow \right )$ Suppose that the real sequences $\{a_n\}$ and $\{b_n\}$ converge to $a$ and $b$ respectively, and consider the complex sequence $\{c_n\} = \{a_n + ib_n\}$ . With $c = a+bi$ , the triangle inequality gives $|c_n - c| = |(a_n + ib_n) - (a+bi)| = |(a_n -a) + i(b_n -b)| \leq |a_n-a|+ |b_n -b|$ Now by the convergence of $\{a_n\}$ and $\{b_n\}$ , for any $\epsilon >0$ there exists an integer $N$ such that $n>N$ implies $|a_n-a| < \frac {\epsilon }{2} \quad \text {and} \quad |b_n -b| < \frac {\epsilon }{2}$ and hence, for $n>N$ we have $|c_n-c| < \frac {\epsilon }{2}+\frac {\epsilon }{2}= \epsilon$ Thus $\{c_n\}$ converges to $c$ .
$\left (\Leftarrow \right )$ Suppose the complex sequence $\{c_n\} = \{a_n + ib_n\}$ converges to $c = a+bi$ . Let $\epsilon >0$ , then since $\{c_n\}$ converges to $c$ , we can choose $N$ so that for $n>N$ , $|c_n - c| < \epsilon$ . Using the fact that $|z| \geq |\Rep z|$ and $|z| \geq |\Imp z|$ , we have $|c_n - c| = |(a_n + ib_n) - (a+bi)| = |(a_n -a) + i(b_n -b)| \geq \max \left \{|a_n-a|, |b_n -b|\right \}$ we can conclude that for $n>N$ $|a_n-a| < \epsilon \quad \text {and} \quad |b_n -b| < \epsilon$ Thus $\{a_n\}$ converges to $a$ and $\{b_n\}$ converges to $b$ .
$\blacksquare$

example 2 Show that the complex sequence $\displaystyle \left \{\tan ^{-1} n + i\sqrt [n]n\right \}$ converges to $\displaystyle \frac {\pi }{2} + i$ .
Since the real sequences $\displaystyle \left \{\tan ^{-1} n\right \}$ and $\displaystyle \left \{\sqrt [n]n\right \}$ converge to $\displaystyle \frac {\pi }{2}$ and $1$ respectively, the complex sequence $\displaystyle \left \{\tan ^{-1} n + i\sqrt [n]n\right \}$ converges to $\displaystyle \frac {\pi }{2} + i$ .

(problem 2) Determine if the following sequences converge or diverge.
a) $\dis \;\; \left \{\frac {n}{n+1}+ i\frac {\cos n}{n} \right \} \quad$ convergesdiverges limit: $\answer {1}$
b) $\dis \;\; \left \{\frac {n^2}{2^n}+ i\sqrt [n]{n^2 +1} \right \} \quad$ convergesdiverges limit: $\answer {i}$
c) $\dis \;\; \left \{\left (1+\frac {1}{n}\right )^n+ i\frac {\ln n}{n} \right \} \quad$ convergesdiverges limit: $\answer {e}$
d) $\dis \;\; \left \{\frac {n^2}{1+n^3}+i\frac {n^2}{1+n} \right \} \quad$ convergesdiverges

Here is a video solution to problem 2b:

2024-09-27 14:07:08