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I'm studying radius of convergence, what happened here? $$\rho = \lim_{n \rightarrow \infty} \left| \frac{\frac{(-1)^n 2^n c_0}{n!}}{\frac{(-1)^{n+1} 2^{n+1} c_0}{n!}} \right|= \lim_{n \rightarrow \infty} \frac{n+1}{2} = \infty$$

Why isnt $\frac{(-1)^n}{(-1)^{n+1}}= \frac{(-1)^n}{(-1)^{n}(-1)^{1}} = \frac{1}{-1}$ ?

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Ok, I see the absolut value bars. How to delete this! –  Noteventhetutorknows Jun 30 '11 at 18:43
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just click on the "delete" link which should be on the bottom left of the post. –  Arturo Magidin Jun 30 '11 at 18:44
    
Oh, isn't the limit 1/2? –  Sunni Jun 30 '11 at 19:01
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You forgot to use $(n+1)!$, instead you used $n!$ twice. –  GEdgar Jun 30 '11 at 19:08
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Suppose you want to find the radius of convergence of the power series $$ \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{2^n }}{{n!}}x^n } . $$ Let $A_n = ( - 1)^n \frac{{2^n }}{{n!}}x^n$, and use the ratio test: $$ \mathop {\lim }\limits_{n \to \infty } \bigg|\frac{{A_{n + 1} }}{{A_n }}\bigg| = \mathop {\lim }\limits_{n \to \infty } \Bigg|\frac{{\frac{{( - 1)^{n + 1} 2^{n + 1} x^{n + 1} }}{{(n + 1)!}}}}{{\frac{{( - 1)^n 2^n x^n }}{{n!}}}}\Bigg| = \mathop {\lim }\limits_{n \to \infty } \bigg|\frac{{ - 2x}}{{n + 1}}\bigg| = 0. $$ From this conclude that the series converges for all $x \in \mathbb{R}$, and hence the radius of convergence is $\infty$.

EDIT: Note that $$ \sum\limits_{n = 0}^\infty {( - 1)^n \frac{{2^n }}{{n!}}x^n } = \sum\limits_{n = 0}^\infty {\frac{{( - 2x)^n }}{{n!}}} = e^{ - 2x} . $$

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