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Find the radius of convergence for the following power series:

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My workings: $$\lim_{n\rightarrow ∞}|\frac{(n+1)! (x-1)^{n+1}}{2^{n+1}(n+1)^{n+1}}\centerdot \frac{2^nn^n}{n!(x-1)^n}|$$

$$=\lim_{n\rightarrow ∞}|\frac{1}{(1+1/n)^n}\centerdot \frac{x-1}{2}|$$

$$=\lim_{n\rightarrow ∞}|\frac{x-1}{2e}|$$

$|x-1|<2e$

$|x|<2e+1$; Therefore, $R=2e+1$?

However, my answer is incorrect, as the answer given is $R=2e$

Could any of you tell me where did I go wrong?

Larry Lee
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  • The domain $|x-1|<2e$ is strictly included in the domain $|x|<2e+1$ hence no, the radius of convergence is not $2e+1$. – Did May 28 '15 at 19:43
  • The interval of convergence is $1-2e\le x\lt 1+2e$. The radius of convergence is $2e$, half the length of the interval of convergence, or equivalently the distance between $1$ and the outer boundary of the interval. – André Nicolas May 28 '15 at 19:45

2 Answers2

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Your calculation is fine, up to the very last step. That is, once, you have $\left | x-1 \right |\leq 2e$, you're done because the series is centered at $x_{0}=1$, so $R=2e$

Matematleta
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Actually, you got the answer right. The disk is just not centered at the origin, since the series is centered at 1.

VJP
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