- #1
alnix
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- 0
Homework Statement
serie ((-1)^(n-1)(x-2)^(n-1))/(5^n)
Homework Equations
how to find the interval of convergence for this?
How to Find the Interval of Convergence for this Series?
- Thread starter alnix
- Start date
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- Convergence Interval
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SUMMARY
The discussion focuses on finding the interval of convergence for the series defined by the formula \((-1)^{n-1}(x-2)^{n-1}/5^n\). The recommended method to determine the radius of convergence is the "ratio test," where the terms \(a_n\) and \(a_{n+1}\) are defined as \(\frac{(-1)^{n-1}(x-2)^{n-1}}{5^n}\) and \(\frac{(-1)^n(x-2)^n}{5^{n+1}}\), respectively. Participants are encouraged to calculate the limit of \(\left|\frac{a_{n+1}}{a_n}\right|\) as \(n\) approaches infinity to identify the values of \(x\) for which this limit is less than 1. The thread emphasizes the importance of showing prior attempts when seeking help.
PREREQUISITES- Understanding of series and convergence concepts
- Familiarity with the ratio test for series convergence
- Basic algebraic manipulation of series terms
- Knowledge of limits in calculus
- Study the application of the ratio test in detail
- Learn about other convergence tests, such as the root test
- Explore the concept of power series and their convergence intervals
- Practice solving various series convergence problems
Students studying calculus, particularly those focusing on series and convergence, as well as educators looking for effective teaching methods for these concepts.
serie ((-1)^(n-1)(x-2)^(n-1))/(5^n)
how to find the interval of convergence for this?
- #2
tiny-tim
Science Advisor
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hi alnix! 
(try using the X2 button just above the Reply box
)
first, simplify it to something you can recognise!
(try using the X2 button just above the Reply box
)first, simplify it to something you can recognise!
- #3
lendav_rott
- 232
- 10
Before we can go anywhere, is this what you mean?
\frac{(-1)^{n-1}(x-2)^{n-1}}{5^n}
\frac{(-1)^{n-1}(x-2)^{n-1}}{5^n}
- #4
HallsofIvy
Science Advisor
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The simplest way to determine the radius of convergence is to use the "ratio test".
a_n= \frac{(-1)^{n-1}(x- 2)^{n-1}}{5^n}
a_{n+1}= \frac{(-1)^n(x- 2)^n}{5^{n+1}}
What is \left|\frac{a_{n+1}}{a_n}\right|? What is the limit of that as n goes to infinity? For what x is that limit less than 1?
a_n= \frac{(-1)^{n-1}(x- 2)^{n-1}}{5^n}
a_{n+1}= \frac{(-1)^n(x- 2)^n}{5^{n+1}}
What is \left|\frac{a_{n+1}}{a_n}\right|? What is the limit of that as n goes to infinity? For what x is that limit less than 1?
- #5
Mark44
Mentor
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To all who have responded in this thread - if someone posts a problem with no effort shown, please use the Report button to let the mentors know.
Alnix, I am closing this thread. Please start a new thread and be sure to show what you have attempted.
Alnix, I am closing this thread. Please start a new thread and be sure to show what you have attempted.
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