Convergence and Continuity of Series: Radius of Convergence R = 1

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SUMMARY

The series \sum \limits_{n=1}^{\infty} \frac{z^{n+1}}{n(n+1)} has a radius of convergence R = 1. It converges for every z in the set \{w \in \mathbb{C} | |w| \leq 1\}. The continuity of the function g(z) = \sum \limits_{n=1}^{\infty} \frac{z^{n+1}}{n(n+1)} is established within the same set. The ratio test confirms the convergence criteria for the series.

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Hummingbird25
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Hi

Looking at the series

[tex]\sum \limit_{n=1} ^{\infty} \frac{z^{n+1}}{n(n+1)}[/tex]

This series has the radius of Convergence R = 1.

Show that the series

converge for every [tex]z \in \{w \in \mathbb{C} | |w| \leq 1 \}[/tex]

And Secondly I need to show that

[tex]g(z) = \sum \limit_{n=1} ^{\infty} \frac{z^{n+1}}{n(n+1)}[/tex]

Is continius in [tex]z \in \{w \in \mathbb{C} | |w| \leq 1 \}[/tex]

Solution:

(1)

Since R = 1, then

[tex]\displaystyle \lim_{n \rightarrow \infty} b_n = \displaystyle \lim_{n \rightarrow \infty} \frac{1}{n(n+1)} = 0[/tex]

[tex]b_n = \displaystyle \lim_{n \rightarrow \infty} b_n = \displaystyle \lim_{n \rightarrow \infty} \frac{1}{(n+1)(n+1)+1} = b_{n +1}[/tex]

Therefore converge the [tex]z \in \{w \in \mathbb{C} | |w| \leq 1 \}[/tex]

(2) Doesn't that follow from (1) ?

Sincerely Yours
Hummingbird25
 
Last edited:
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no, use the ratio test to determine R.
 
Applying the ratio to the original series

I get

[tex]\displaystyle \lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| \\<br /> = \lim_{n \rightarrow \infty} |\frac{z^{n+2}}{(n+1)(n+2)} * \frac {n(n+1)}{z^{n+1}}| = |z| < 1[/tex]

Then do I use a specific test show that the original series ?

converge for every z \in \{w \in \mathbb{C} | |w| \leq 1 \

and the is Is continius in [tex]z \in \{w \in \mathbb{C} | |w| \leq 1 \}[/tex]

Sincerely Yours
Hummingbird25

benorin said:
no, use the ratio test to determine R.
 

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