Calculating the interval and radius of convergence

Question

I have $$s_n=\frac{-6^n(x+8)^n}{\sqrt{n}}$$ and I have to find the interval and radius of convergence.

Using ratio test:

$$\frac{-6^{n+1}(x+8)^{n+1}}{\sqrt{n+1}}\cdot \frac{\sqrt{n}}{-6^n(x+8)^n}$$

$$\lim_{n\rightarrow \infty}-6(x+8)\frac{\sqrt{n}}{\sqrt{n+1}}$$

The last term goes to 1.

Since the interval must be smaller than 1, we get:

$$-6(x+8)<1$$

This gives $$x<-\frac{43}{6}$$, which should be the interval of convergence. Half of that again is the radius of converge, so $$x<-\frac{43}{12}$$

But this is wrong,

any ideas where the error lies and what the right solution is?

Thanks

Answer 1

The correct typing for $s_n$ seems to be $$s_n=\frac{(-6)^n(x+8)^n}{\sqrt n}.$$

Compute the same ratio test but for $|s_n|$ (not $s_n$ which takes negative values).

Eventually, you get that your series is absolutely convergent if $$6|x+8|<1,$$ i.e. $$x\in\left(-8-\frac16,-8+\frac16\right),$$ and trivially divergent if $x\notin\left[-8-\frac16,-8+\frac16\right].$

The radius of convergence $R$ is half the length of the interval of convergence $\left(-8-\frac16,-8+\frac16\right)$:$$R=\frac16.$$

If $x=-8-\frac16$, $\sum_{n=1}^\infty s_n=\sum_{n=1}^\infty\frac1{\sqrt n}$ is divergent.

If $x=-8+\frac16$, $\sum_{n=1}^\infty s_n=\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt n}$ is a conditionally convergent alternating series.