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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?

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This is a very standard, very straightforward problem; what have you tried? – Brian M. Scott Nov 8 '12 at 16:59
Use the Ratio Test. Informally, the problem is that $n!$ grows hugely fast. – André Nicolas Nov 8 '12 at 17:01
Try by using the ratio test – M. Strochyk Nov 8 '12 at 17:02
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@everyone except the OP: – The Chaz 2.0 Nov 8 '12 at 17:40

2 Answers 2

up vote 0 down vote accepted

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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