4.2 Complex Series

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We determine convergence of complex series.

A complex series is defined as a limit of its sequence partial sums: $\sum _{k=1}^\infty c_k = \lim _{n \to \infty } \sum _{k=1}^n c_k = \lim _{n \to \infty } S_n$ If the sequence $\{S_n\}$ converges, then we say the series converges. Otherwise, we say that the series diverges.

Test for Divergence If $\dis \sum _{k=1}^\infty c_k$ converges, then $\dis \lim _{k\to \infty } c_k = 0$

Proof: Note that $c_n = S_n - S_{n-1}$ , the difference in partial sums. Since the sequence $\{S_n\}$ converges, given $\epsilon >0$ , there exists a number $N$ and a number $S$ such that if $n>N$ then $|S_n - S| < \epsilon /2$ . Hence for $n > N+1$ , the triangle inequality gives \begin{align*} |c_n| &= |S_n - S_{n-1}| = |(S_n -S) + (S-S_{n-1})| \\ &\leq |S_n -S| + |S-S_{n-1}| < \frac{\epsilon }{2} + \frac{\epsilon }{2} =\epsilon \end{align*}
Hence, the sequence $\{c_n\}$ converges to zero. $\blacksquare$

example 1 Determine whether the complex series $\dis \sum _{k=1}^\infty \frac {1+ik}{k}$ converges.
Since, $\dis \lim _{k \to \infty } \frac {1+ik}{k} = i$ . Since $i \neq 0$ , the Test for Divergence says that the series $\dis \sum _{k=1}^\infty \frac {1+ik}{k}$ does not converge, i.e., it diverges.

(problem 1) Determine whether the complex series $\dis \sum _{k=0}^\infty \frac {k!}{(2i)^k}$ converges.
Since, $\dis \lim _{k \to \infty } \left |\frac {k!}{(2i)^k}\right | = \answer {\infty }$ ,
$\dis \lim _{k \to \infty } \frac {k!}{(2i)^k} \neq \answer {0}$ ,
the series $\dis \sum _{k=0}^\infty \frac {k!}{(2i)^k}\;\;$ convergesdiverges
by the Test for ConvergenceTest for Divergence

Here is a video solution to problem 1:

Geometric Series Let $c \in \C$ and consider the series $\sum _{k=0}^\infty c^k = 1 + c + c^2 + c^3 + \cdots$ This series converges if and only if $|c| < 1$ . Moreover, if it converges, its sum is $\displaystyle \frac {1}{1-c}$ .

The complex number $c$ in the geometric series $\dis \sum _{k=0}^\infty c^k$ is called the common ratio of the series.

Proof: Suppose $|c| < 1$ . Then the partial sums have a closed form whose limit can be computed directly. Since $S_n = \sum _{k=1}^n c^k = 1+c+c^2+\cdots + c^n$ we have $cS_n = c+c^2+\cdots + c^{n+1}$ Subtracting gives $S_n - cS_n = 1-c^{n+1}$ from which we obtain the closed form $S_n = \frac {1-c^{n+1}}{1-c}$ Now, since $|c|<1$ , we have $c^n \to 0$ as $n \to \infty$ and hence $\sum _{k=0}^\infty c^k = \lim _{n\to \infty } S_n = \lim _{n\to \infty } \frac {1-c^{n+1}}{1-c} = \frac {1}{1-c}$ Thus, the geometric series converges and its sum is also known.
If $|c| \geq 1$ then $c^n \nrightarrow 0$ as $n \to \infty$ which implies that the geometric series diverges (by the Test for Divergence). $\blacksquare$

example 2 Determine whether the geometric series converges or diverges:
a) $\;\;\dis \sum _{k=0}^\infty \frac {1}{(1+i)^k}\quad$ b) $\;\;\dis \sum _{k=0}^\infty \left (\frac {1+i}{1-i}\right )^k$
Part a) This is a geometric series with common ratio $\dis \frac {1}{1+i}$ . The modulus of the common ratio is $\left |\frac {1}{1+i}\right | = \frac {1}{\sqrt 2} <1$ Hence the geometric series converges and its sum is $\sum _{k=0}^\infty \frac {1}{(1+i)^k} = \frac {1}{1-\frac {1}{1+i}} = \frac {1+i}{i} = 1-i$ Part b) This is a geometric series with common ratio $\dis \frac {1+i}{1-i}$ . The modulus of the common ratio is $\left |\frac {1+i}{1-i}\right | = \frac {\sqrt 2}{\sqrt 2} = 1 \nless 1$ Hence the geometric series diverges.

(problem 2) Determine whether the geometric series converges or diverges. If it converges, find its sum: $\;\;\dis \sum _{k=0}^\infty \left (\frac {2i}{2+i}\right )^k$
The common ratio is $\; \answer {\frac {2i}{2+i}}$
The modulus of the common ratio is $\; \answer {\frac {2}{\sqrt 5}}$
Is this less than 1? yesno
The geometric series convergesdiverges
The sum of the series is $\; \answer {\frac {2+i}{2-i}}$

Here is a video solution to problem 2:

Absolute Convergence If the series $\dis \sum _{k=1}^\infty |c_k|$ converges, then so does the complex series $\dis \sum _{k=1}^\infty c_k$ .

To prove this result, we will need the Monotone Convergence Theorem which states that a bounded, monotone sequence of real numbers converges.

Proof: Suppose $\displaystyle \sum _{k=1}^\infty |c_k|$ converges to the real number $S$ , and break the partial sums of the complex series $\displaystyle \sum _{k=1}^\infty c_k$ into their real and imaginary parts: $\sum _{k=1}^n c_k = \sum _{k=1}^n a_k + i\sum _{k=1}^n b_k$ where $a_k, b_k \in \R$ for all $k$ . Let $a_k^+ = a_k$ if $a_k \geq 0$ and let $a_k^+ = 0$ if $a_k < 0$ . Similarly, let $a_k^- = -a_k$ if $a_k < 0$ and let $a_k^- = 0$ if $a_k \geq 0$ . Then $\sum _{k=1}^n a_k = \sum _{k=1}^n a_k^+ - \sum _{k=1}^n a_k^- = A_n^+ - A_n^-$ The sequences $\{A_n^+\}$ and $\{A_n^-\}$ are both increasing sequences and since $|a_k| \leq |c_k|$ for all $k$ , they are both bounded above by $S$ . Hence, by the Monotone Convergence Theorem, both sequences converge. Thus, the real series $\dis \sum _{k=1}^n a_k$ converges, to a sum, $A$ . Applying the same argument to the partial sums of the imaginary part, $\dis \sum _{k=1}^n b_k$ , we see that it too converges to a sum, $B$ . Finally, we can conclude that $\dis \sum _{k=1}^\infty c_k$ converges to $A + iB$ . $\blacksquare$

example 3 Show that the complex series converges: $\dis \sum _{k=1}^\infty \frac {2e^i}{(k+ik)^2}$
The modulus of $c_k$ is $\left |\frac {2e^i}{(k+ik)^2}\right | = \frac {2}{2k^2} = \frac {1}{k^2}$ and the series $\sum _{k=1}^\infty \frac {1}{k^2}$ is a convergent $p$ -series with $p = 2 > 1$ . Moreover the sum of this well known series is $\frac {\pi ^2}{6}$ .

(problem 3) Determine whether the complex series $\dis \sum _{k=0}^\infty \frac {i^k}{k!}$ converges.
The modulus of $c_k$ is $\; \answer {1/k!}$
The series $\dis \sum _{k=0}^\infty \frac {1}{k!}\;\;$ convergesdiverges
The series $\dis \sum _{k=0}^\infty \frac {i^k}{k!}\;\;$ convergesdiverges

Here is a video solution to problem 3:

We define the limit superior of a sequence of real numbers $\{a_n\}, \; \limsup _{n \to \infty } a_n,\;$ to be the largest limit of any subsequence of $\{a_n\}$ .

If the sequence $\{a_n\}$ converges to $a$ , then $\limsup _{n \to \infty } a_n =a$ .

It is possible that $\limsup _{n \to \infty } a_n =\pm \infty$

example 4 Find the limit superior of the sequence $\{c_k\}= \{1, 2, 3, 1, 2, 3, 1, 2, 3, \ldots \}$
The sequence $\{c_k\}$ has subsequences that converge to $1, 2$ and $3$ . Since the largest of these limits is $3$ , we have $\limsup _{k\to \infty } c_k = 3$

(problem 4) Find the limit superior of the sequence:
a) $\dis \; \{(-1)^k\} \quad \limsup _{k\to \infty } (-1)^k = \; \answer {1}$
b) $\dis \; \{c_k\} = \{1, -1, \frac 12, -1, \frac 13, -1, \ldots \} \quad \limsup _{k\to \infty } c_k = \; \answer {0}$

Here is a video solution to problem 4:

Root Test Let $\displaystyle \sum _{k=1}^\infty c_k$ be a complex series and let $L = \limsup _{k \to \infty } \sqrt [k]{|c_k|}$ Then the series converges if $L <1$ and diverges if $L >1$ .

Proof: First, suppose $\dis L = \limsup _{k \to \infty } \sqrt [k]{|c_k|} < 1$ . Then there exists a natural number $N$ such that for $k>N |c_k| < L^k$ . Hence the series $\dis \sum _{k = N+1}^\infty |c_k|$ converges by direct comparison to the geometric series $\dis \sum _{k = N+1}^\infty L^k$ . Thus, the series $\dis \sum _{k = 1}^\infty c_k$ converges absolutely and is therefore convergent.
Next, suppose $\dis L = \limsup _{k \to \infty } \sqrt [k]{|c_k|} > 1$ Then there exists a subsequence of $\{c_k\}$ such that $\sqrt [k]{|c_k|} > 1$ which implies that $|c_k| > 1$ for each of the infinitely many terms in this subsequence. Hence, $c_k \nrightarrow 0$ as $k \to \infty$ . By the Test for Divergence, the series $\displaystyle \sum _{k=1}^\infty c_k$ diverges. $\blacksquare$

The series $\dis \sum _{k=1}^\infty \frac {1}{k^2}$ and $\dis \sum _{k=1}^\infty \frac {1}{k}$ show that if $L = 1$ in the Root Test, then no conclusion can be drawn. This is because the first series converges and the second series diverges but $L=1$ in both cases.

example 5 Determine if the series converges or diverges: $\dis \sum _{k=1}^\infty \left (\frac {k+i}{k^2 - i}\right )^k$
Computing the $k^{th}$ root of the modulus of $c_k$ gives $\sqrt [k]{|c_k|} = \sqrt [k]{\left |\left (\frac {k+i}{k^2 - i}\right )^k\right |} = \sqrt [k]{\left |\frac {k+i}{k^2 - i}\right |^k} = \sqrt {\frac {k^2 + 1}{k^4 + 1}}$ and the limit superior is $\limsup _{k \to \infty } \sqrt [k]{|c_k|} = \limsup _{k \to \infty } \sqrt {\frac {k^2 + 1}{k^4 + 1}} = 0 <1$ so the series converges.

(problem 5) Determine if the series converges or diverges: $\sum _{k=1}^\infty \left (\frac {e^{ik}}{1+i\sqrt k}\right )^k$ $\sqrt [k]{|c_k|} = \; \answer {\frac {1}{\sqrt {1+k}}}$
The limit superior is $\; \limsup _{k\to \infty }\sqrt [k]{|c_k|}= \; \answer {0}$
Is this less than 1? yesno
The series convergesdiverges

Here is a video solution to problem 5:

Ratio Test Let $\displaystyle \sum _{k=1}^\infty c_k$ be a complex series and let $L = \limsup _{k \to \infty } \left |\frac {c_{k+1}}{c_k}\right |$ Then the series converges if $L <1$ and diverges if $L >1$ .

If $c_k = 0$ for infinitely many values of $k$ , then the ratio test cannot be used.

The proof of the ratio test is similar to the proof of the root test.

example 6 Determine if the series converges or diverges: $\sum _{k=0}^\infty \frac {i^k2^k}{k!}$ The modulus of the ratio of successive terms is $\left |\frac {i^{k+1}2^{k+1}}{(k+1)!}\cdot \frac {k!}{i^k 2^k}\right | = \frac {2}{k+1}$ Taking the limit gives $\limsup _{k \to \infty } \frac {2}{k+1} = 0 < 1$ By the Ratio Test, the series converges.

(problem 6) Determine if the series converges or diverges: $\sum _{k=0}^\infty \frac {2e^{ik}}{(2k)!}$ $\left |\frac {c_{k+1}}{c_k}\right | = \; \answer {\frac {1}{(2k+2)(2k+1)}}$
The limit superior is $\; \limsup _{k\to \infty }\left |\frac {c_{k+1}}{c_k}\right |= \; \answer {0}$
Is this less than 1? yesno
The series convergesdiverges

Here is a video solution to problem 6:

2024-11-04 22:54:30