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Chapter 4: Problem 7

In the geometric series, show that if \(|z|>1\), then \(\lim _{n \rightarrow \infty}\left|S_{n}\right|=\infty\).

Short Answer

Expert verified
If \(|z| > 1\), then \(\lim_{n\to\infty} |S_n| = \infty\).

Step by step solution

01

Understanding the Geometric Series Sum

In a geometric series, the sum of the first \( n \) terms, \( S_n \), is given by \( S_n = a + ar + ar^2 + \, \cdots \, + ar^{n-1} \). This can also be represented as \( S_n = a \frac{r^n - 1}{r - 1} \) if \( r eq 1 \).
02

Express Basic Series in Terms of Single Variable

For simplicity in proof, consider the series with first term \( a = 1 \) and common ratio \( r = z \). Thus, the sum \( S_n = \frac{z^n - 1}{z - 1} \).
03

Take Absolute Value of the Sum

Consider the absolute value of the sum: \( |S_n| = \left| \frac{z^n - 1}{z - 1} \right| \). This reflects the magnitude of \( S_n \) in the complex plane.
04

Evaluate Conditions for \(|z| > 1\)

Assume \(|z| > 1\), implying that as \( n \to \infty \), \(|z^n| \to \infty \). Since the constant \(|z - 1|\) does not grow as \( n \to \infty \), it can be considered a non-dominating factor.
05

Limit Calculation for \(|S_n|\)

Given \( |z| > 1 \), the factor \( |z^n| \) grows without bound, causing \( |S_n| = \left| \frac{z^n - 1}{z - 1} \right| \approx \frac{|z^n|}{|z - 1|} \) to also grow towards infinity as \( n \to \infty \).
06

State the Conclusion

Therefore, as \( n \to \infty \), \(|S_n| \to \infty\) for \(|z| > 1\), proving the result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. It is a foundational concept in mathematics that helps us understand series and sequences. A simple form of a geometric series is the sum of the first few terms.
  • For a geometric series with the first term as 1 and common ratio \( r \), the sum of the first \( n \) terms is represented as \( S_n = \frac{r^n - 1}{r - 1} \) when \( r eq 1 \).
  • This formula allows us to quickly determine the sum up to any number of terms.
If the common ratio \( r \) is greater than 1, each subsequent term becomes larger. As a result, the series sum grows significantly as more terms are included. This property is crucial when evaluating limits in the context of complex numbers.
Complex Plane
The complex plane is a way to visualize complex numbers geometrically. It is an invaluable tool in complex analysis. Each complex number corresponds to a point on the plane.
  • The horizontal axis represents the real part of the complex number.
  • The vertical axis represents the imaginary part.
The magnitude or modulus of a complex number \( z = x + yi \) is found using \( |z| = \sqrt{x^2 + y^2} \). This represents the distance from the point \( z \) to the origin.
In our original exercise, the condition \( |z| > 1 \) helps determine the direction and limit behavior of the series in the complex plane. For a complex number with a large modulus, the series grows quickly, displaying rapidly increasing distance from the origin with each term added.
Limit Calculation
Limit calculation is a technique to determine what happens to a sequence or function as its input approaches some value, often infinity. In complex analysis, limits help us analyze the behavior of series like geometric series.
  • For a geometric series \( S_n = \frac{z^n - 1}{z - 1} \), the limit as \( n \to \infty \) allows us to predict the sum's behavior as more terms are added.
  • In this exercise, if \( |z| > 1 \), then \( |z^n| \to \infty \). As such, \( |S_n| = \frac{|z^n|}{|z - 1|} \) also approaches infinity since \(|z-1|\), a constant factor, does not grow with \( n \).
Learning limit calculations will broaden one's ability to solve a myriad of problems, especially those involving converging or diverging series in mathematics.

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Most popular questions from this chapter

Suppose that \(\left\\{z_{0}, z_{1}\right\\}\) is a 2 -cycle for \(f\). (a) Show that if \(z_{0}\) is attracting for \(g_{2}(z)\), then so is the point \(z_{1}\). Hint: Differentiate \(g_{2}(z)=f(f(z))\), using the chain rule, and show that \(g_{2}^{\prime}\left(z_{0}\right)=g_{2}^{\prime}\left(z_{1}\right) .\) (b) Generalize part (a) to \(n\) -cycles.

Let \(\sum_{k=1}^{\infty}\left(x_{n}+i y_{n}\right)=u+i v .\) If \(c=a+i b\) is a complex constant, show that \(\sum_{n=1}^{\infty}(a+i b)\left(x_{n}+i y_{n}\right)=(a+i b)(u+i v)\)

Prove that \(\lim _{n \rightarrow \infty} z_{n}=0\) iff \(\lim _{n \rightarrow \infty}\left|z_{n}\right|=0 .\)

Prove that a sequence can have only one limit. Hint: Suppose that there is a sequence \(\left\\{z_{n}\right\\}\) such that \(z_{n} \rightarrow \zeta_{1}\) and \(z_{n} \rightarrow \zeta_{2}\). Show this implies \(\zeta_{1}=\zeta_{2}\) by proving that for all \(\varepsilon>0,\left|\zeta_{1}-\zeta_{2}\right|<\varepsilon\).

Prove that Newton's method always works for polynomials of degree 1 (functions of the form \(f(z)=a z+b\), where \(a \neq 0\) ). How many iterations are necessary before Newton's method produces the solution \(z=-\frac{1}{a}\) to \(f(z)=0 ?\)
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