Power Series: Interval Of Convergence

In summary, the individual is having trouble with a series and is specifically struggling with the squared aspect of the first question. They are unsure of how to approach it and also have doubts about the second question. The answer for the first question is 4 (-4,4) and the answer for the second question is infinity (-infinity,+infinity). The individual ends up realizing that they made a mistake and the correct answer for the second question is actually infinity. They also receive clarification on how to solve the first question and provide a step-by-step explanation.
  • #1
tak13
8
0

Homework Statement



I am not really good with Series so I having a hard time with these problems.

http://img835.imageshack.us/img835/858/img1257d.jpg [Broken]

Homework Equations


The Attempt at a Solution



The part I am stuck is where I highlighted. The first question: The whole thing is squared so I don't know how to do it with the squared in the way.

The second question: It is just plain weird. I think I am not supposed to do that for the highlighted part.

The answer for first question is : 4 (-4,4)
The answer for second question is : infinity (-infinity,+infinity)
 
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  • #2
First:

[tex]\lim_{n\to \infty} \left| \frac{\left(n \cdot (n-1) \cdot (n-2) \ldots \right) ^{2} \cdot (2n+2)(2n+1)}{\left( (n+1) \cdot n \cdot (n-1) \cdot (n-2) \ldots \right) ^{2}} \right| = \lim_{n\to \infty} \left| \frac{2(n+1)(2n+1)}{(n+1)^{2}} \right| = \lim_{n\to \infty}\left| \frac{2(2n+1)}{(n+1)} \right| = 4[/tex]

On the second one you're basically done.

[tex]\lim_{n\to \infty} \left| \frac{(2n+2)(2n+1)}{(2n+3)} \right| = \infty \ .[/tex]

Think the answers are mixed up.
 
  • #3
Haha, I just looked over the manual, the second question does seem to be like that. I guess I messed up looking at the manual the first time.

For the first one do you mean I should do like this:

http://img694.imageshack.us/img694/4568/img1259u.jpg [Broken]
 
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  • #4
tak13 said:
Haha, I just looked over the manual, the second question does seem to be like that. I guess I messed up looking at the manual the first time.

For the first one do you mean I should do like this:

Yep. Factor the numerator and you'll get

[tex]\lim_{n\to \infty} \left| \frac{2(n+1)(2n+1)}{(n+1)^2} \right| = \lim_{n\to \infty} \left| \frac{2(2n+1)}{n+1} \right|[/tex]

which evaluates to 4 (divide numerator and denominator with [tex]n[/tex]).
 

What is a power series?

A power series is an infinite series of the form n=0^∞ an(x-c)n, where an are the coefficients and c is the center of the series.

What is the interval of convergence?

The interval of convergence is the interval of values for x for which the power series converges. It can be either a closed interval, an open interval, or a single value.

How do you determine the interval of convergence?

The interval of convergence can be determined by using the ratio test or the root test. Both of these tests involve taking the limit as n goes to infinity of the absolute value of the ratio or root of consecutive terms in the series. If the limit is less than 1, then the series converges. If the limit is greater than 1, the series diverges. If the limit is equal to 1, further tests are needed to determine convergence or divergence.

What is the radius of convergence?

The radius of convergence is the distance from the center of the power series to the nearest point where the series converges. It is represented by the letter R and can be calculated using the ratio or root test.

What happens if the value of x is outside the interval of convergence?

If the value of x is outside the interval of convergence, then the power series will diverge and not give a meaningful value. It is important to note that the endpoints of the interval may or may not be included in the convergence, so they should be tested separately.

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