Find the radius of convergence and convergence interval

Question

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 ?

Answer 1

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 $[ -17, 1)$.

Answer 2

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.

Answer 3

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.