Power Series (Complex Analysis)

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September 21st 2008, 07:04 PM
shadow_2145

Power Series (Complex Analysis)

Find the radius of convergence of the given power series.

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

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

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

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

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

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

Any help would be appreciated, thanks!
September 21st 2008, 07:14 PM
jbpellerin

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
September 21st 2008, 10:40 PM
choovuck

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

(2) denote $z^3=t$ . Then we have the series $\sum_{j=1}^{\infty}t^j/2^j$ , so use (b):
$\lim a_n^{1/n}=\lim 1/2=1/2$ , so the series converge for any $|t|<2$ . This means the original series converge for any $|z|=|t|^{1/3}<2^{1/3}$ , i.e. $R=2^{1/3}$
September 22nd 2008, 09:20 AM
Moo

Quote:

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... :/
September 22nd 2008, 05:54 PM
choovuck

Oh, I think initially he wrote $1/R=\lim \frac{|a_n|}{|a_{n+1}|}$ rather than $1/R=\lim \frac{|a_{n+1}|}{|a_{n}|}$ . So either he edited his post after my reply, or I just got confused. in any case, the answer is right.
September 25th 2008, 03:07 PM
shadow_2145

Quote:

Originally Posted by choovuck

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

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

Why does $R = 4$ instead of $\frac{1}{4}$ in problem (1)?
September 25th 2008, 03:23 PM
shadow_2145

I have a new problem I am stuck with too.

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

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

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

Right path so far? Way off? Help will be appreciated, thanks!
September 25th 2008, 05:06 PM
choovuck

Quote:

Originally Posted by shadow_2145

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

because 1/R is limit of a_{n+1}/a_n, so R is limit of a_n/a_{n+1}
September 25th 2008, 05:07 PM
choovuck

Quote:

Originally Posted by shadow_2145

I have a new problem I am stuck with too.

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

I don't understand if you mean $5^{-n}$ or $5^{(-1)^n}$
September 25th 2008, 06:13 PM
shadow_2145

Quote:

Originally Posted by choovuck

I don't understand if you mean $5^{-n}$ or $5^{(-1)^n}$

Oops, my bad. $5^{(-1)^n}$ is the correct one.
September 25th 2008, 06:14 PM
shadow_2145

Quote:

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!
September 25th 2008, 06:39 PM
choovuck

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