Finding the Centre and Radius of Convergence of Power Series: Explained

[MissP.25_5]

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?

 

[Quesadilla]

A power series expansion is an infinite series
\begin{equation*}
\sum_{n=0}^{\infty} a_n (z-c)^n,
\end{equation*}
where ##a_n## are coefficients and ##c \in \mathbb{C}## is the center (the point we are expanding about).

I think you can figure out why
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{4i(z-i)(n+1)}{n+3} = 4i(z-i).
\end{equation*}
The easiest might be to substitute ##k = f(n)##, for some well-chosen function ##f##.

 

[HallsofIvy]

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)

This is incorrect. The "ratio test" only applies to series of non-negative real numbers. In order to apply the ratio test to more general series, you must take the absolute value
[tex]\lim{m\to\infty} \frac{4|z- i|(n+1)}{n+ 3}= 4|z- i|\lim_{n\to\infty} \frac{n+1}{n+ 3}[/tex].

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

Well, what do you get when you "divide all term" of [itex]\frac{n+1}{n+ 3}[/itex] by n?

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

Why does the centre become i?

The "ratio" test says that series [tex]\sum a_n[/tex] converges if [tex]\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1[/tex]. What is [tex]\lim_{\n to\infty}4|z- i|\frac{n+1}{n+ 3}[/tex]? For what values of z is that less than 1?

 

[MissP.25_5]

Well, what do you get when you "divide all term" of [itex]\frac{n+1}{n+ 3}[/itex] by n?


The "ratio" test says that series [tex]\sum a_n[/tex] converges if [tex]\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1[/tex]. What is [tex]\lim_{\n to\infty}4|z- i|\frac{n+1}{n+ 3}[/tex]? For what values of z is that less than 1?

The answer to your first question is 1 and for the second, z≤i. Right?

 

[HallsofIvy]

"[itex]z\le i[/itex]" doesn't even make sense. The complex numbers are NOT an "ordered" field. You keep forgetting the absolute value!