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

[Hummingbird25]

Hi

Looking at the series

\sum \limit_{n=1} ^{\infty} \frac{z^{n+1}}{n(n+1)}

This series has the radius of Convergence R = 1.

Show that the series

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

And Secondly I need to show that

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

Is continius in z \in \{w \in \mathbb{C} | |w| \leq 1 \}

Solution:

(1)

Since R = 1, then

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

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

Therefore converge the z \in \{w \in \mathbb{C} | |w| \leq 1 \}

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

Sincerely Yours
Hummingbird25

 

[benorin]

no, use the ratio test to determine R.

 

[Hummingbird25]

Applying the ratio to the original series

I get

\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| &lt; 1

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 z \in \{w \in \mathbb{C} | |w| \leq 1 \}

Sincerely Yours
Hummingbird25

no, use the ratio test to determine R.