Determine the radius of convergence of the power series $\sum \limits _{n=4} ^\infty \frac {2n+4} {4^{n+5}} (x-8)^{4n+1}$.

I tried the ratio test to find where $\frac {a_n} {a_{n+1}} < 1$ but I ended up with $\frac {(2n+6)(x-8)^4} {4(2n+4)} < 1$ and I don't know where to go from there.

  • 2
    $\begingroup$ You need to take the limit of this expression as $n\to\infty$. $\endgroup$ – user84413 May 26 '15 at 22:35

$$\lim_{n\to\infty} f(x) \frac {(2n+6)(x-8)^4} {4(2n+4)} < 1$$

$\iff$ $$\lim_{n\to\infty} f(x)(x-8)^4 < \frac{4(2n+4)}{(2n+6)}$$

$\iff$ $$\lim_{n\to\infty} f(x)(x-8)^4 < \frac{4 + 8/n}{1+3/n}$$

$\iff$ $(x-8)^4 $< 4


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