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Chapter 9: Problem 6

Determine whether the given series must diverge because its terms do not converge to \(0 .\) $$ \sum_{n=1}^{\infty}\left(\frac{-1}{7}\right)^{n} $$

Short Answer

Expert verified
The series does not diverge because its terms converge to 0.

Step by step solution

01

Understand the Series

The given series is \( \sum_{n=1}^{\infty}\left(\frac{-1}{7}\right)^{n} \). This is a geometric series with common ratio \( r = \frac{-1}{7} \).
02

Identify the Conditions for Convergence

A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if the absolute value of the common ratio \( |r| < 1 \).
03

Check the Term for Convergence

For the series to converge, the terms of the series need to approach \( 0 \) as \( n \to \infty \). In a geometric series, this means \( (\frac{-1}{7})^n \to 0 \) as \( n \to \infty \).
04

Evaluate the Limit of the Terms

Since the absolute value \( |\frac{-1}{7}| = \frac{1}{7} < 1 \), the terms \( (\frac{-1}{7})^n \) do indeed converge to \( 0 \).
05

Conclusion on Divergence

Since the terms converge to \( 0 \), the series \( \sum_{n=1}^{\infty}\left(\frac{-1}{7}\right)^{n} \) does not diverge merely because of its term limit. The terms meet the convergence condition of a geometric series because \(|r| < 1\).

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Key Concepts

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

Convergence of a Series
A series converges if the sequence of its partial sums approaches a finite limit. In simpler terms, if you keep adding the terms of the series together, the total gets closer and closer to a specific number.
For a geometric series, which has the form \[ \sum_{n=0}^{\infty} ar^n \] to converge, the absolute value of the common ratio \( |r| \) must be less than 1.
Here is why this rule helps us determine convergence:
  • If \(|r| < 1\), the terms \( r^n \) become very small as \( n \) becomes very large. Hence, the series approaches a definite sum.
  • If \(|r|\) is greater than 1 or equal to 1, the terms do not get smaller, and the series grows without bound.
For the series \( \sum_{n=1}^{\infty}\left(\frac{-1}{7}\right)^{n} \), the common ratio is \( \frac{-1}{7} \). Since \( \left|\frac{-1}{7}\right| = \frac{1}{7} < 1 \), this geometric series converges.
Infinite Series
An infinite series is a sum with infinitely many terms. Imagine you keep adding terms forever without stopping. Although it seems counterintuitive, some infinite series do converge to a specific limit.
Key characteristics of infinite series:
  • They have a defined mathematical limit, even though you are adding infinitely many numbers.
  • The partial sums, which are sums of the first \(n\) terms, help us understand if a series converges or not.
When dealing with infinite series, not all of them will converge. It depends on the nature of the terms being added. In the case of our series, \( \sum_{n=1}^{\infty}\left(\frac{-1}{7}\right)^{n} \), knowing that the terms grow smaller due to the absolute value of the ratio being less than 1, helps us predict that there is a finite sum that these terms are approaching.
Divergence in Series
Divergence occurs when the terms of a series, added up, do not approach any finite limit. When a series diverges, its partial sums increase indefinitely, meaning they keep getting larger and never settle on a specific value.
Reasons why a series might diverge:
  • If the terms do not tend toward zero as you go further along in the series.
  • If the absolute value of the common ratio \( |r| \) in a geometric series is greater than or equal to 1.
For our example, the series \( \sum_{n=1}^{\infty}\left(\frac{-1}{7}\right)^{n} \), the terms \( \left(\frac{-1}{7}\right)^{n} \) converge to zero as \( n \) increases, because \( \left|\frac{-1}{7}\right| < 1 \). Therefore, the series does not diverge based on its term's behavior.

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Most popular questions from this chapter

a. Show that \(\ln \frac{1}{1-x}=\sum_{n=1}^{\infty} \frac{x^{n}}{n}\). b. Using (a), show that \(\ln 2=\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\). c. Using the fact that $$ \sum_{n=N}^{\infty} \frac{1}{n 2^{n}} \leq \sum_{n=N}^{\infty} \frac{1}{N 2^{n}} $$ estimate \(\ln 2\) with an error less than \(0.01\).

Find the Taylor series of the given function about \(a\). Use the series already obtained in the text or in previous exercises. $$ f(x)=10^{x} ; a=0 $$

Determine which series diverge, which converge conditionally, and which converge absolutely. $$ \sum_{n=3}^{\infty}(-1)^{n+1} \frac{1}{n(n-2)} $$

Approximate the sum of the given series with an error less than \(0.001\). $$ \sum_{n=2}^{\infty}(-1)^{n+1} \frac{1}{1+n+6 n^{2}} $$

When a particle of mass \(m\) moves with a high velocity \(v\), the theory of relativity implies that its kinetic energy is given by $$ K=m c^{2}\left(\frac{1}{\sqrt{1-v^{2} / c^{2}}}-1\right) $$ where \(c\) is the speed of light. Using (2), show that when the ratio \(\mathrm{v} / \mathrm{c}\) is small, then \(K\) is approximately equal to the usual "Newtonian" kinetic energy \(\frac{1}{2} m v^{2}\). (Thus the relativistic kinetic energy reduces to the Newtonian kinetic energy when the velocity is small.)
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