# Thread: Power Series (Complex Analysis)

1. ## Power Series (Complex Analysis)

Find the radius of convergence of the given power series.

(1) $\displaystyle {\sum_{k=0}}^{\infty}$ $\displaystyle \frac{(k!)^2}{(2k)!}(z-2)^k$

(2) $\displaystyle {\sum_{j=0}}^{\infty}$ $\displaystyle \frac{z^{3j}}{2^j}$

I'm not sure how to use the theorems given in my book to solve these.

Theorem: Suppose that $\displaystyle {\sum_0}^{\infty}$$\displaystyle a_n(z-z_0)^n$ is a power series with a positive or infinite radius of convergence R.

(a) If $\displaystyle lim_{n \to\infty} |a_{n+1}/a_n|$ exists, then $\displaystyle \frac{1}{R} = lim_{n \to\infty}|\frac{a_{n+1}}{a_n}|$

(b) If $\displaystyle lim_{n\to\infty}\sqrt[n]{|a_n|}$ exists, then $\displaystyle \frac{1}{R} =lim_{n\to\infty}\sqrt[n]{|a_n|}$

Any help would be appreciated, thanks!

2. try the ratio test
so a(n+1)/a(n) = L
and the series is only convergent when L < 1

so if you can come up with an inequality and then solve for z

3. (1) use ratio test:
$\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\lim\frac{(2n +2)(2n+1)}{(n+1)^2}=4$, so $\displaystyle R=4$ (I think you have the mistake in the theorem in part (a))

(2) denote $\displaystyle z^3=t$. Then we have the series $\displaystyle \sum_{j=1}^{\infty}t^j/2^j$, so use (b):
$\displaystyle \lim a_n^{1/n}=\lim 1/2=1/2$, so the series converge for any $\displaystyle |t|<2$. This means the original series converge for any $\displaystyle |z|=|t|^{1/3}<2^{1/3}$, i.e. $\displaystyle R=2^{1/3}$

4. Originally Posted by choovuck
(I think you have the mistake in the theorem in part (a))
And why ?
Isn't it equivalent to what you have done ? In any book talking about it, I've seen this rule stated this way... :/

5. Oh, I think initially he wrote $\displaystyle 1/R=\lim \frac{|a_n|}{|a_{n+1}|}$ rather than $\displaystyle 1/R=\lim \frac{|a_{n+1}|}{|a_{n}|}$. So either he edited his post after my reply, or I just got confused. in any case, the answer is right.

6. Originally Posted by choovuck
(1) use ratio test:
$\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\lim\frac{(2n +2)(2n+1)}{(n+1)^2}=4$, so $\displaystyle R=4$ (I think you have the mistake in the theorem in part (a))

(2) denote $\displaystyle z^3=t$. Then we have the series $\displaystyle \sum_{j=1}^{\infty}t^j/2^j$, so use (b):
$\displaystyle \lim a_n^{1/n}=\lim 1/2=1/2$, so the series converge for any $\displaystyle |t|<2$. This means the original series converge for any $\displaystyle |z|=|t|^{1/3}<2^{1/3}$, i.e. $\displaystyle R=2^{1/3}$

Why does $\displaystyle R = 4$ instead of $\displaystyle \frac{1}{4}$ in problem (1)?

7. I have a new problem I am stuck with too.

$\displaystyle \sum_{n = 0}^{\infty} 5^{(-1)n}z^n$

Seems simple enough, but I suck at power series...

$\displaystyle \frac{1}{R} = lim_{n \to \infty} \sqrt[n]{5^{(-1)}} = lim_{n \to \infty} (\sqrt[n]{5})^{(-1)}$

Right path so far? Way off? Help will be appreciated, thanks!

Why does $\displaystyle R = 4$ instead of $\displaystyle \frac{1}{4}$ in problem (1)?
because 1/R is limit of a_{n+1}/a_n, so R is limit of a_n/a_{n+1}

I have a new problem I am stuck with too.

$\displaystyle \sum_{n = 0}^{\infty} 5^{(-1)n}z^n$
I don't understand if you mean $\displaystyle 5^{-n}$ or $\displaystyle 5^{(-1)^n}$

10. Originally Posted by choovuck
I don't understand if you mean $\displaystyle 5^{-n}$ or $\displaystyle 5^{(-1)^n}$
Oops, my bad. $\displaystyle 5^{(-1)^n}$ is the correct one.

11. Originally Posted by choovuck
because 1/R is limit of a_{n+1}/a_n, so R is limit of a_n/a_{n+1}

Oh, I see now. Hehe, duh!

12. $\displaystyle 5^{(-1)^n}$ means that it's the sequence 1/5, 5, 1/5, 5, ..., and then $\displaystyle \lim_{n\to\infty} \sqrt[n]{a_n}=1$ (since both $\displaystyle \lim_{n\to\infty} \sqrt[n]{5}=1$ and $\displaystyle \lim_{n\to\infty} \sqrt[n]{1/5}=1$), so R=1