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Let be given the Laurent series:

i) for $z \ne -1$ and $z_{0}=-1$, $$ (z-3)\sin\left(\frac{1}{z+1}\right)= \displaystyle{ 4 \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!(z+1)^{2n+1}} + \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!(z+1)^{2n}}} $$ ii) for $z\ne 1,2$ and $z_{0}=1$, $$ \frac{z}{(z-1)(z-2)} =-2\sum_{n=0}^{\infty} (z-1)^{n} - \frac{1}{z-1} $$ How to find their radii of convergence?

my attemptions:

i) Substitute $x^{-1}:= z+1$, this gives $$ 4 \sum \frac{(-1)^{n+1} (x)^{2n+1}}{(2n+1)!} + \sum \frac{(-1)^{n}(x)^{2n}}{(2n+1)!} $$ Then the radius of convergence for $\sum \frac{(-1)^{n+1} (x)^{2n+1}}{(2n+1)!}$ is $\lim_{n\rightarrow \infty} |\frac{a_{n}}{a_{n+1}}| = \lim |\frac{(2n+3)!}{(2n+1)!}| = \infty$ so for $|x|< \infty \Leftrightarrow |\frac{1}{z+1}|<\infty \Rightarrow |z|< 0 $ and the radius of convergence for $\sum \frac{(-1)^{n}(x)^{2n}}{(2n+1)!}$ is also 0 .

ii) The left side is a taylor series with $R=1$ and I don't know what to do with the right side!

attempt 2 :

i) find $a_{n} $ and $a_{n-}$: $a_{n}$ is zero in i) and $a_{n-} = a_{1n-}+a_{2n-}$ consists of $a_{1n-} = \frac{(-1)^{n+1}}{(2n+1)!(z+1)^{2n+1}}$ and $a_{2n-} = \frac{(-1)^{n}}{(2n+1)!}$

ii) $a_{n}= 1 $ and $a_{n-}=1$ but thats not possible…

attempt 3 :

i) $1<|z|<\infty$

ii) $1<|z|<2$

is this correct?

Somebody sees the right way? Tell me. Please.

share|improve this question
    
That $|\frac1{z+1}|$ finite implies $|z|<0$ is, well... creative. –  Did Nov 6 '11 at 14:16
    
Then what other way to find out radius of convergence for laurent series? –  VVV Nov 6 '11 at 14:39
1  
I suggest you begin by addressing the content of my comment. Although this comment is brief and expressed somewhat lightly, I believe it hints at a deep, incredibly basic, misunderstanding on your part. Unless you come clear on this specifically, I see no point whatsoever in going further. Hence: What does the condition that $|\frac1{z+1}|$ is finite mean? –  Did Nov 6 '11 at 14:45
    
$|\frac{1}{z+1}| < \infty $ implies that $1 < \infty$ which is true for all z so $R=\infty$ ? ? ? –  VVV Nov 6 '11 at 14:50
1  
@Ross, yes. As is explained competently in the first paragraph of the WP page I linked to, five comments ago... :-) –  Did Nov 6 '11 at 16:24

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