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Find the radius of convergence of the given power series:

$\sum _{n=0}^{ \infty} \frac{8n!x^n}{2^n}$

After taking the limits as n-> $\infty$, I get $\frac{8x}{2}$, and Radius of convergence is R = 2. Is this correct?

  • 0
    Try by using the [ratio test](http://en.wikipedia.org/wiki/Ratio_test)2012-11-08

2 Answers 2

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Using the Ratio Test, it looks like we get $R=0$.

$\left|a_{n+1}\over a_{n}\right| = \left| \frac{8(n+1)! \cdot x^{n+1} \cdot 2^n}{2^{n+1} \cdot 8n! \cdot x^n} \right| = \left| \frac{8(n+1)n! \cdot x \cdot x^n \cdot 2^n}{2\cdot 2^n \cdot 8n! \cdot x^n} \right|$

$=\frac{\left|x\right|} {2} \lim_{n\to\infty} (n+1) \to \infty \ \ \forall x \ne 0\implies R=0$

Thus, the power series only converges when $x=c$, which is at $x=0$ here.

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This can help you solve this problem.

The answer is:

$0$