Problem 5 - Arbeitsbuch Mathematik für Ingenieure

Question

Untersuchen Sie die folgenden unendlichen Reihen auf Konvergenz und bestimmen Sie gegebenfalls die Summe (außer bei f)). Entscheiden Sie bei den Reihen c) bis \(\mathrm{f}\) ), für welche \(z \in \mathbb{C}\) die Reihe jeweils konvergiert. a) \(\sum_{k=0}^{\infty}(1-i)^{k}\) b) \(\sum_{k=1}^{\infty}\left(\frac{1-i}{2}\right)^{k}\) c) \(\sum_{k=0}^{\infty}\left(\frac{z+1}{z-1}\right)^{k}\) d) \(\sum_{k=0}^{\infty}(-1)^{k} \frac{z^{2 k}}{k !}\) e) \(\sum_{k=1}^{\infty}\left(\frac{4+3 i}{5 z}\right)^{k}\) f) \(\sum_{k=1}^{\infty} \frac{z^{n}}{n^{2} \cdot 2^{n}}\)

Step-by-Step Solution

Answer

a) Divergent, b) Convergent, sum = i, c) Convergent if \(|z+1| < |z-1|\), d) Convergent, sum = \(e^{-z^2}\), e) Convergent if \(|z| > 1\), f) Convergent for all \(z\).

Step 1: Analyze Series a

Consider the series \(\sum_{k=0}^{\infty}(1-i)^{k}\). This is a geometric series with first term \(a = 1\) and common ratio \(r = (1-i)\). A geometric series converges if and only if \(|r| < 1\). Here, \( r = (1-i)\), so \(|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} > 1\). Since \(|r| > 1\), the series diverges.

Step 2: Analyze Series b

Consider the series \(\sum_{k=1}^{\infty}\left(\frac{1-i}{2}\right)^{k}\). This is a geometric series with first term \(a = \frac{1-i}{2}\) and common ratio \(r = \frac{1-i}{2}\). A geometric series converges if and only if \(|r| < 1\). Here, \(|\frac{1-i}{2}| = \frac{\sqrt{1^2 + (-1)^2}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} < 1\). Thus, the series converges. The sum of the series is \(S = \frac{a}{1-r} = \frac{\frac{1-i}{2}}{1-\frac{1-i}{2}} = \frac{1-i}{1+i} = i\).

Step 3: Analyze Series c

Consider the series \(\sum_{k=0}^{\infty}\left(\frac{z+1}{z-1}\right)^{k}\). This is a geometric series with common ratio \(r = \frac{z+1}{z-1}\). A geometric series converges if and only if \(|r| < 1\). Thus, for the series to converge, \(|\frac{z+1}{z-1}| < 1\). Therefore, the series converges for \(z eq 1\) and \(|z+1| < |z-1|\).

Step 4: Analyze Series d

Consider the series \(\sum_{k=0}^{\infty}(-1)^{k} \frac{z^{2 k}}{k!}\). This series can be recognized as the Maclaurin series for the exponential function \(e^x\) evaluated at \(x=-z^2\). Thus, the series converges for all \(z \in \mathbb{C}\) and sums to \(e^{-z^2}\).

Step 5: Analyze Series e

Consider the series \(\sum_{k=1}^{\infty}\left(\frac{4+3i}{5z}\right)^{k}\). This is a geometric series with common ratio \(r = \frac{4+3i}{5z}\). A geometric series converges if and only if \(|r| < 1\). Thus, the series converges for \(|\frac{4+3i}{5z}| < 1\). Since \(|4+3i| = 5\), this simplifies to \(|z| > 1\).

Step 6: Analyze Series f

Consider the series \(\sum_{k=1}^{\infty} \frac{z^n}{n^2 \cdot 2^n}\). Since the general term \(\frac{z^n}{n^2 \cdot 2^n} = \frac{1}{n^2} \cdot \left(\frac{z}{2}\right)^n \) involves \( \left(\frac{z}{2}\right)^n\) and the series \(\sum_{k=1}^{\infty} \frac{1}{n^2}\) converges (p-series with p > 1), this series converges for all \(z \in \mathbb{C}\).

Key Concepts

geometric series

In mathematics, a geometric series is a series with a constant ratio between successive terms. If the first term of the series is denoted by \(a\) and the common ratio by \(r\), the geometric series is represented as \(\sum_{k=0}^{\infty} ar^k\).
The convergence of a geometric series is determined by the value of the common ratio.
Specifically, a geometric series converges if and only if the absolute value of the common ratio is less than one, \(|r| < 1\).
If the series converges, its sum can be calculated using the formula: \ \[ S = \frac{a}{1-r} \].
For example, consider the series \(\sum_{k=1}^{\infty} \left( \frac{1-i}{2} \right)^k\). Here, the first term \( a = \frac{1-i}{2} \) and the common ratio \( r = \frac{1-i}{2} \). To check convergence, we calculate the absolute value of \( r \): \ \[ \left| \frac{1-i}{2} \right| = \frac{\sqrt{1^2 + (-1)^2}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} < 1 \]. Since \( \left| r \right| < 1 \), the series converges. The sum of the series can thus be found using the sum formula: \ \[ S = \frac{a}{1-r} = \frac{\frac{1-i}{2}}{1-\frac{1-i}{2}} = \frac{1-i}{1+i} = i \].

complex numbers

Complex numbers are numbers that have both a real part and an imaginary part and are typically written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\).
The set of complex numbers is denoted by \(\mathbb{C}\).
In the context of series, complex numbers often appear in the common ratio of geometric series or as parameters in other series.
For example, consider the series \(\sum_{k=0}^{\infty}(1-i)^k\). Here, the common ratio \(r = 1-i\), which is a complex number.
To determine convergence, we compute the magnitude (or modulus) of the common ratio: \ \[ |1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} > 1 \].
Since the magnitude is greater than 1, the series diverges. Complex numbers also play a crucial role in determining regions of convergence for series involving complex parameters.
For example, in the geometric series \(\sum_{k=0}^{\infty}\left(\frac{z+1}{z-1}\right)^k\), convergence depends on the value of the complex parameter \(z\). The series converges if the magnitude of the ratio \(\frac{z+1}{z-1}\) is less than 1:
\ \[ \left| \frac{z+1}{z-1} \right| < 1 \].

Maclaurin series

The Maclaurin series is a special case of the Taylor series, centered at zero. It provides a way to represent functions as infinite sums of powers of the variable, which is particularly useful for approximations.
The Maclaurin series for a function \( f(x) \) is given by: \ \[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k \].
This is especially useful when dealing with functions like the exponential function. For instance, the Maclaurin series for \(e^x\) is: \ \[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \]. Using this concept, we can recognize series forms that match this pattern.
Consider the series \(\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k}}{k!}\). This can be identified as the Maclaurin series for \(e^x\) evaluated at \(x = -z^2\). \ \[ \sum_{k=0}^{\infty}(-1)^k \frac{z^{2k}}{k!} = e^{-z^2} \].
Hence, this series converges for all \(z \in \mathbb{C}\), providing a succinct way to determine both the convergence and the sum.

exponential function

The exponential function \(e^x\) is one of the most important functions in mathematics, characterized by the fact that its derivative is equal to itself.
It is defined by the infinite series: \ \[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \].
This series converges for all real and complex values of \(x\).
Exponential functions often appear in series and integrals.
For example, consider the series \(\sum_{k=0}^{\infty}(-1)^k \frac{z^{2k}}{k!}\), which is the Maclaurin series for the exponential function evaluated at \( x = -z^2 \).
This means that the series converges for all complex numbers \(z\) and sums to \(e^{-z^2}\).
The exponential function's properties make it a powerful tool for solving differential equations, modeling growth and decay processes, and exhibiting complex behavior in various mathematical contexts.

p-series

A \( p \)-series is a series of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), where \(p\) is a real number.
The convergence of a \( p \)-series depends on the value of \( p \).
Specifically, a \( p \)-series converges if and only if \( p > 1 \). For example, the series \(\sum_{k=1}^{\infty} \frac{1}{n^2}\) converges because \( p = 2 > 1 \).
Consider a series like \(\sum_{k=1}^{\infty} \frac{z^n}{n^2 \cdot 2^n}\). This involves a geometric component \(\left( \frac{z}{2} \right)^n \) and a \( p \)-series component \( \frac{1}{n^2} \).
The geometric series \(\sum_{k=0}^{\infty} \left( \frac{z}{2} \right)^n \) converges by itself when \( | \frac{z}{2} | < 1 \). Combined with the \( p \)-series which converges for \( p = 2 \), we conclude that the overall series converges for all \(z \in \mathbb{C}\), offering a powerful demonstration of how these series types intertwine to determine convergence.