Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 6
\(3-8=\) Test the series for convergence or divergence. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{n}{\sqrt{n^{3}+2}} $$
Short Answer
Expert verified
The series converges by the Alternating Series Test.
Step by step solution
01
Identify Alternating Series
This given series \( \sum_{n=1}^{\infty}(-1)^{n} \frac{n}{\sqrt{n^{3}+2}} \) is an alternating series because of the factor \((-1)^{n}\), which causes the terms to alternate between positive and negative values.
02
Apply Alternating Series Test (AST)
According to the Alternating Series Test, the series \( \sum (-1)^{n} a_n \) converges if the absolute value \( a_n \) of the terms is a positive decreasing sequence that approaches zero. Let \( a_n = \frac{n}{\sqrt{n^{3}+2}} \). We need to show that \( a_n \to 0 \) as \( n \to \infty \).
03
Check if Terms Approach Zero
Evaluate the limit of \( a_n \) as \( n \to \infty \):\[ \lim_{n \to \infty} \frac{n}{\sqrt{n^{3}+2}}\]To simplify, we divide the numerator and the denominator by \( n^{3/2} \):\[ \lim_{n \to \infty} \frac{n}{n^{3/2}} \cdot \frac{1}{\sqrt{1 + \frac{2}{n^3}}} = \lim_{n \to \infty} \frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{1 + \frac{2}{n^3}}}\]The limit of \( \frac{1}{\sqrt{n}} \) is zero as \( n \to \infty \), so \( a_n \to 0 \).
04
Check if Sequence is Decreasing
To verify if \( a_n = \frac{n}{\sqrt{n^{3}+2}} \) is decreasing, check that \( a_{n+1} < a_n \):We need:\[\frac{n+1}{\sqrt{(n+1)^3 + 2}} < \frac{n}{\sqrt{n^{3}+2}}\]By inspection for large \(n\), the denominator grows faster than the numerator sees, ensuring that \( a_n \) decreases as \( n \) increases.
05
Conclusion of Alternating Series Test
Since \( a_n \to 0 \) and \( a_n \) is decreasing, by the Alternating Series Test, the series converges.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Convergence and Divergence
In calculus, understanding the convergence and divergence of a series is fundamentally important. A series converges if the sum of its infinite terms approaches a specific finite value. Conversely, a series diverges if it doesn't approach any particular value, potentially growing indefinitely or oscillating without settling. For the given series in the exercise, determining its convergence involves applying a test designed to handle alternating series, which switch signs between terms. This test helps ascertain if the series will ultimately sum to a finite number, or if it will continue to fluctuate endlessly without finding a home at a specific numerical value. If the test confirms convergence, it suggests that the series behaves in a manner similar to convergent geometric or arithmetic series and will result in a definitive sum. If it diverges, it behaves more like a row of dominos that never settle into place, forever in motion without rest.
Infinite Series
An infinite series is essentially the sum of an infinite sequence of numbers. It is a core concept in mathematical analysis that can represent complex constructs in a more comprehensible form. Each term in an infinite series signifies a contribution towards this potentially infinite summation. The alternating nature of the series given in the exercise, made evident by the \((-1)^{n}\) factor, means the terms switch between positive and negative values. This switch is critical in determining how closely and in what manner the series approaches a limit, if it does at all. The goal when analyzing an infinite series is to identify whether the cumulative effects of its terms will result in convergence or divergence. In simpler terms, we want to know if the infinite sum trends towards a particular value or if it remains unbounded.
Calculus Series Tests
Calculus provides several tests for determining the behavior of series, each suited for different kinds of series. The Alternating Series Test (AST) is one specific tool, ideal for series where the terms alternate in sign. For the exercise's series, AST helps us see if the positive part of each term decreases steadily and approaches zero, a necessary condition to conclude convergence for alternating series.
To apply this test, you first make sure that the terms become negligibly small as their index, or position, within the sequence increases. This reduction means you need to check two things: the limit of the sequence approaching zero and the term sequence being decreasing. If both conditions are satisfied, as in the exercise provided, the AST confirms convergence.
Understanding the right test to use, such as AST, can often simplify complex series evaluations into more manageable parts, serving as a reliable pathway to understanding the infinite intricacies of calculus.
To apply this test, you first make sure that the terms become negligibly small as their index, or position, within the sequence increases. This reduction means you need to check two things: the limit of the sequence approaching zero and the term sequence being decreasing. If both conditions are satisfied, as in the exercise provided, the AST confirms convergence.
Understanding the right test to use, such as AST, can often simplify complex series evaluations into more manageable parts, serving as a reliable pathway to understanding the infinite intricacies of calculus.
