/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A computer program gives the res... [FREE SOLUTION] | 91Ó°ÊÓ

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A computer program gives the result \(1 / 6\) for the sum of the series \(\sum_{n=0}^{\infty}(-5)^{n} .\) Show that this series is divergent. Do you see what happened? Warning hint: Always consider whether an answer is reasonable, whether it's a computer answer or your work by hand.

Short Answer

Expert verified
The series \(\sum_{n=0}^{\infty}(-5)^{n}\) is divergent because \(\lim_{n \to \infty} (-5)^n\) does not approach zero.

Step by step solution

01

Define the Series

The given series is \(\textstyle \sum_{n=0}^{\infty}(-5)^{n}\).
02

Identify the General Term

The general term of the series is \(a_n = (-5)^n\).
03

Apply the Divergence Test

To determine if the series \(\textstyle \sum a_n\) is convergent or divergent, check the limit of the general term \(a_n\) as \(n\) approaches infinity. \(\textstyle \lim_{n \to \infty}a_n = \lim_{n \to \infty} (-5)^n \).
04

Evaluate the Limit

Since \(-5\) is a constant and not within the interval \(-1, 1\), \((-5)^n\) grows exponentially as \(n\) increases. Hence, \(\lim_{n \to \infty} (-5)^n = \infty\) or \(-\big.\infty\), depending on whether \(n\) is even or odd.
05

Conclusion from Divergence Test

Since \(\lim_{n \to \infty} a_n \) does not approach zero, the series \(\textstyle \sum_{n=0}^{\infty}(-5)^{n}\) is divergent.
06

Analyze the Given Result

The result given by the computer program, \(\frac{1}{6}\), is not reasonable because it violates the divergence test results. The series \(\textstyle \sum_{n=0}^{\infty}(-5)^{n}\) is divergent and should not sum to any finite value.

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Key Concepts

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

Divergence Test
One of the core strategies in determining whether an infinite series converges or diverges is the Divergence Test. This method hinges on a simple yet essential concept: If the limit of the general term of the series as it approaches infinity does not equal zero, the series cannot possibly converge.
For a series \(\textstyle \sum_{n=0}^{\infty} a_n\) to be convergent, it's necessary (but not sufficient) that \(\textstyle \lim_{n \to \infty}a_n = 0\). If this limit is not zero, the series must diverge.
Let's apply it to our example: the series \(\textstyle \sum_{n=0}^{\infty}(-5)^{n}\). We look at the general term \((a_n = (-5)^n)\). As n approaches infinity, the term \((-5)^n\) grows exponentially, and does not tend to zero. Hence, the series \(\textstyle \sum_{n=0}^{\infty}(-5)^{n}\) diverges.
Infinite Series
Understanding infinite series is key in calculus and beyond. An infinite series is the sum of the terms of an infinite sequence. It's written in the form \(\textstyle \sum_{n=0}^{\infty} a_n\). Each term in the series follows a specific pattern or rule.
Infinite series are captivating for their properties and their wide applications in mathematics, physics, and engineering. They can either converge to a specific value or diverge, implying they don’t sum up to a finite result.
Convergence and divergence of series are essential aspects to study. For instance, if a series converges, its partial sums \(\textstyle S_n = \sum_{k=0}^{n}a_k\) approach a finite value as n approaches infinity. On the flip side, if these partial sums do not approach a finite value, the series is said to diverge.
Exponential Growth
Exponential growth is a concept where quantities increase rapidly over time. It holds special significance in infinite series and is crucial for understanding why certain series diverge.
In mathematics, exponential growth refers to sequences where the factor of growth is consistently multiplied, such as the term \((-5)^n\). As n increases, this term grows exponentially.
When dealing with the series \(\textstyle \sum_{n=0}^{\infty}(-5)^{n}\), each additional term becomes exponentially larger since \((-5)^n\) vastly increases in magnitude as n increases. This exponential growth ensures that \((-5)^n\) doesn't settle towards zero, leading to the divergence of the series.
  • Exponential growth is characterized by rapid escalation, often depicted as a swiftly rising curve.
  • This rapid increase explains the infinite nature of many divergent series such as the one in the exercise.

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Most popular questions from this chapter

(a) Using computer or tables (or see Chapter 7 , Section 11 ), verify that \(\sum_{n=1}^{\infty} 1 / n^{2}=\) \(\pi^{2} / 6=1.6449+,\) and also verify that the error in approximating the sum of the series by the first five terms is approximately 0.1813 . (b) \(\quad\) By computer or tables verify that \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)(1 / 2)^{n}=\pi^{2} / 12-(1 / 2)(\ln 2)^{2}=\) \(0.5822+,\) and that the sum of the first five terms is \(0.5815+.\) (c) Prove theorem (14.4). Hint: The error is \(\left|\sum_{N+1}^{\infty} a_{n} x^{n}\right| .\) Use the fact that the absolute value of a sum is less than or equal to the sum of the absolute values. Then use the fact that \(\left|a_{n+1}\right| \leq\left|a_{n}\right|\) to replace all \(a_{n}\) by \(a_{N+1},\) and write the appropriate inequality. Sum the geometric series to get the result.

Find the Maclaurin series for the following functions. $$\cos [\ln (1+x)]$$

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{n !}{(2 n) !}$$

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{3^{n}}$$

Show that if \(p\) is a positive integer, then \(\left(\begin{array}{l}p \\\ n\end{array}\right)=0\) when \(n>p,\) so \((1+x)^{p}=\sum\left(\begin{array}{l}p \\\ n\end{array}\right) x^{n}\) is just a sum of \(p+1\) terms, from \(n=0\) to \(n=p .\) For example, \((1+x)^{2}\) has 3 terms, \((1+x)^{3}\) has 4 terms, etc. This is just the familiar binomial theorem.

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