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Given $$S=\sum^{\infty}_{0}\frac{2^n(x-2)^n}{(n+2)!}$$

After using root test, I got, $$-1\le\frac{2(x-2)}{(n+2)}\le 1$$ The n did not cancel. Now, How do I conclude about radius of convergence?

Thank You.

I did mistake, I used ratio test.

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You need to take the limit $n\to\infty$ in order to apply the test. – user12477 Mar 21 '13 at 10:11
It seems you applied the ratio test, not the root test. Which is fine: take the limit as suggested. – GEdgar Mar 21 '13 at 12:40
yes. i did so. and got solution. thank you. – user45099 Mar 21 '13 at 12:47

Hint: Radius of convergence= $1/\lim \sup_{n\to \infty }|a_n|^{\frac{1}{n}}$ when the power series is $\sum_{i=0}^{\infty}a_nx^n$.

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up vote 2 down vote accepted

The ratio test seems more suitable to me in this case: $$ \lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1\implies\sum\limits_{n=0}^\infty a_n\text{ converges absolutely}$$ So in this case we have $$\lim\limits_{n\to\infty}\left|\frac{2^{n+1}(x-2)^{n+1}}{(n+1+2)!}\frac{(n+2)!}{2^n(x-2)^n}\right|=\lim\limits_{n\to\infty} \frac{2(x-2)}{n+3}=0<1\quad \forall x\in \mathbb R$$ So the radius of convergence is .... (you should be able to fill it in now)

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