Radius of Convergence for Power Series: Mathboy20

[mathboy20]

Hi

I'm told that the the power series:

\sum_{n=0} ^ \infty (2n+1) z^n has the radius of convergens

R = 1.

Proof:

Using the Definition of convergens for power series:

\frac{(2n+1)}{(2n+1)+1} = \frac{(2n+1)}{(2n+3)}

limit _{n \rightarrow \infty} \frac{(2n+1)}{(2n+3)} = 1

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:

\sum_{n = 0} ^{\infty} a_n z^n

that a_n \rightarrow \infty for n \rightarrow \infty

Do I the use this fact here to show that a_n diverges ??

Best Regards
Mathboy20

 

[Jameson]

\frac{(2n+1)}{(2n+1)+1} = \frac{(2n+1)}{(2n+3)}

I'm not following this step. I would use the ratio test for finding the radius of convergence.

 

[kam.epi]

I think this would be the proper way to evaluate the series:

\sum_{n=0} ^ \infty (2n+1) z^n

so using the ratio test,

limit _{n \rightarrow \infty} \frac{(2(n+1)+1) z^(n+1)}{(2n+1) z^n}

limit _{n \rightarrow \infty} \frac{(2n+3)}{(2n+1)}|z|

Therefore, |z|<1

implies that the interval of convergence is -1<z<1

Therefore, the radius of convergence must be 1.

 

[mathboy20]

Hello and thank Your for Your answer,

Anyway if I then have show that the series diverges for all point on the circle of convergens.

Doesn't that mean that

\frac{2n+3}{2n+1}|z| \geq 1, where n \neq 0

if n = 1

then |z| \geq \frac{3}{5}

Am I on the right track here?

Best Regards
Mathboy20

I think this would be the proper way to evaluate the series:

\sum_{n=0} ^ \infty (2n+1) z^n

so using the ratio test,

limit _{n \rightarrow \infty} \frac{(2(n+1)+1) z^(n+1)}{(2n+1) z^n}

limit _{n \rightarrow \infty} \frac{(2n+3)}{(2n+1)}|z|

Therefore, |z|<1

implies that the interval of convergence is -1<z<1

Therefore, the radius of convergence must be 1.

 

[HallsofIvy]

Hello and thank Your for Your answer,

Anyway if I then have show that the series diverges for all point on the circle of convergens.

Doesn't that mean that

\frac{2n+3}{2n+1}|z| \geq 1, where n \neq 0

if n = 1

then |z| \geq \frac{3}{5}

Am I on the right track here?

Best Regards
Mathboy20

n= 1 is not relevant. What is
lim_{n\rightarrow \infty}\frac{2n+3}{2n+1}?