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Power Series: Interval Of Convergence

  • Thread starter tak13
  • Start date
  • #1
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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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Answers and Replies

  • #2
13
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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.
 
  • #4
13
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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]).
 

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