I have two problems in which I'm stuck finding the radius and interval of convergence:
1) $\sum\limits_{n=1}^\infty\frac{n^3(x+4)^n}{4^nn^{11/3}}$
Applying the ratio rule allows me to simply as such:
$(\frac{(n+1)^3(x+4)^{n+1}}{4^{n+1}(n+1)^{11/3}})(\frac{4^nn^{11/3}}{n^3(x+4)^n})$ $= \frac{(n+1)^3n^{11/3}}{4(n+1)^{11/3}n^3}(x+4)$
Since $\lim_{n\to \infty}\frac{(n+1)^3n^{11/3}}{4(n+1)^{11/3}n^3} = \frac{1}{4}$, I get R = $\frac{1}{4}|x+4|< 1$, so the radius of convergence should be 4, with the interval of convergence being (-8, 0).
But apparently either my radius or interval is wrong (or both, I don't know), and I have trouble figuring out why.
2)$\sum\limits_{n=1}^\infty\frac{n^2(x-14)^n}{4*8*12...(4n)}$
And simplifying the power series I get this: $\sum\limits_{n=1}^\infty\frac{n^2(x-14)^n}{4^nn!}$
So once again I apply the ratio rule, which allows me simply as so:
$(\frac{(n+1)^2(x-14)^{n+1}}{4^{n+1}(n+1)!})(\frac{4^nn!}{n^2(x-14)^n})$
= $\frac{(n+1)^2(x-14)}{4(n+1)n^2} = \frac{(n+1)}{4n^2}(x-14)$
Since $\lim_{n\to \infty}\frac{n+1}{4n^2} = 0$, the radius of convergence should also be 0, and the interval should be {14}.
But like 1), either the radius or interval or both is wrong, and I can't figure out why.
I'd appreciate if someone could point me in the right direction. Thanks in advance for all your help!
