Find the radius of convergence and convergence interval Find the radius of convergence and convergence  interval
$$\sum^\infty_{n=1}\frac{n^5\;(x+8)^n}{9^n\; n^\frac{17}{3}}$$
My attempt:
I'm computing $\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|$ and
I got the answer $|9(x+8)|$. So the radius  of convergence $R=\frac{1}{9}$ and interval =$(\frac{-73}{9},\frac{-71}{9})\,$.
Am I right ?
 A: The basic term,
ignoring the shift by 8,
is
$(x/9)^n$.
Comparing this to
$z^n$,
which has radius of convergence
(roc) 1,
the original converges when
$|x/9| < 1$
or
$|x| < 9$.
Therefore its
roc is 9
(not 1/9),
so the interval is
$(-8-9, -8+9)$
or
$(-17, 1)$.
What happens at the endpoints
does depend on
the power of $n$
involved.
A: Probably too complex as an answer but too long for a comment.
$$u_n=\frac{n^5\;(x+8)^n}{9^n\; n^\frac{17}{3}}=\frac{9^{-n} (x+8)^n}{n^{2/3}}$$ Defining $y=\frac {x+8} 9$ makes $$u_n=\frac{y^n}{n^{2/3}}\implies S=\sum_{n=1}^\infty u_n=\text{Li}_{\frac{2}{3}}(y)$$ where appears the polylogarithm function  which has a branch cut discontinuity in the complex  plane running from $1$ to $\infty$ (see here). It is defined as long as $|y|<1$.
In fact, it is even convergent for $y=-1$ since $\text{Li}_k(-1)=-\left(1-2^{1-k}\right) \zeta (k)$.
So, back to $x$, $S$ is convergent for $-17\leq x <1$ as marty cohen answered.
A: The series is a power series around $x = -8$, that is, it is of the form
$$ \sum_{n=1}^{\infty} a_n (x - (-8))^n $$
where $a_n = \frac{n^5}{9^n n^{\frac{17}{3}}} = \frac{1}{9^n n^{\frac{2}{3}}}$. The radius of convergence is given by $R = \lim_{n \to \infty} \frac{a_n}{a_{n+1}}$ (if it exists) and not by $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$! In our case, we have
$$ \frac{a_{n+1}}{a_n} = \frac{9^n n^{\frac{2}{3}}}{9^{n+1}(n+1)^{\frac{2}{3}}} = \frac{1}{9} \left( \frac{n}{n+1} \right)^{\frac{2}{3}}  \to \frac{1}{9} $$
and so the radius of convergence is $R = 9$ and you are guaranteed that the series will converge in $(-17, 1)$. Now you also need to check the end points of the interval to determine the full domain of convergence. 
At $x = 1$ we get the series
$$ \sum_{n=1}^{\infty} \frac{1}{n^{\frac{2}{3}}} $$
which diverges by the $p$ criterion. 
At $x = -17$ we get the series
$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{\frac{2}{3}}} $$
which converges using the alternate series test. Thus, the full interval of convergence is $[ -17, 1)$.
