What is the centre and radius of convergence for a power series?

[aruwin]

Hello.
I need someone to explain to me how to find the centre and radius of convergence of power series.
I got the working and the answers but there are some things I don't understand.

$$\sum_{n=0}^{\infty}\frac{(4i)^n(z-i)^n}{(n+1)(n+2)}$$

Using the ratio test, we got
$$\lim_{{n}\to{\infty}} \frac{4i(z-i)(n+1)}{n+3}$$= $4i(z-i)$

Ok, in this part, why is the limit $4i(z-i)$? Don't we have to divide all the terms by n?

And the final answer is: $R=1/4, z=i$

Why does the centre become i?

 

[chisigma]

Hello.
I need someone to explain to me how to find the centre and radius of convergence of power series.
I got the working and the answers but there are some things I don't understand.

$$\sum_{n=0}^{\infty}\frac{(4i)^n(z-i)^n}{(n+1)(n+2)}$$

Using the ratio test, we got
$$\lim_{{n}\to{\infty}} \frac{4i(z-i)(n+1)}{n+3}$$= $4i(z-i)$

Ok, in this part, why is the limit $4i(z-i)$? Don't we have to divide all the terms by n?

And the final answer is: $R=1/4, z=i$

Why does the centre become i?

The series $\displaystyle \sum_{n=0}^{\infty} \frac{s^{n}}{(n+1)\ (n+2)}$ converges for |s|<1, so that is R=1 and $s_{0} = 0$. Setting $\displaystyle s = 4\ i\ (z-i)$ You obtain that the series $\displaystyle \sum_{n=0}^{\infty} \frac{\{4\ i\ (z-i)\}^{n}}{(n+1)\ (n+2)}$ converges for $\displaystyle |4\ i\ (z-i)|<1$ so that is $R=\frac{1}{4}$ and $z_{0}=i$...

Kind regards

$\chi$ $\sigma$