- #1
mathboy20
- 30
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Hi
I'm told that the the power series:
[tex]\sum_{n=0} ^ \infty (2n+1) z^n[/tex] has the radius of convergens
R = 1.
Proof:
Using the Definition of convergens for power series:
[tex]\frac{(2n+1)}{(2n+1)+1} = \frac{(2n+1)}{(2n+3)} [/tex]
[tex]limit _{n \rightarrow \infty} \frac{(2n+1)}{(2n+3)} = 1[/tex]
Therefore the radius of convergens is R = 1. Right ?
Second question: The Power series above suposedly diverges on every point on the circle of convergens. How do I show that?
I know that according to the definition of divergens of the power series:
[tex]\sum_{n = 0} ^{\infty} a_n z^n [/tex]
that [tex]a_n \rightarrow \infty [/tex] for [tex]n \rightarrow \infty [/tex]
Do I the use this fact here to show that a_n diverges ??
Best Regards
Mathboy20
I'm told that the the power series:
[tex]\sum_{n=0} ^ \infty (2n+1) z^n[/tex] has the radius of convergens
R = 1.
Proof:
Using the Definition of convergens for power series:
[tex]\frac{(2n+1)}{(2n+1)+1} = \frac{(2n+1)}{(2n+3)} [/tex]
[tex]limit _{n \rightarrow \infty} \frac{(2n+1)}{(2n+3)} = 1[/tex]
Therefore the radius of convergens is R = 1. Right ?
Second question: The Power series above suposedly diverges on every point on the circle of convergens. How do I show that?
I know that according to the definition of divergens of the power series:
[tex]\sum_{n = 0} ^{\infty} a_n z^n [/tex]
that [tex]a_n \rightarrow \infty [/tex] for [tex]n \rightarrow \infty [/tex]
Do I the use this fact here to show that a_n diverges ??
Best Regards
Mathboy20
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