Does this $\left(-\frac{\sqrt{7}}{3}\right)^{-n}$ converge to $0$?

Question

$\left(-\frac{\sqrt{7}}{3}\right)^{-n}$

Does this sequence converge to $0$ by the ratio test? Using the ratio test I get $-\frac{\sqrt{7}}{3} \lt 1$

I tried to double check my results in wolfram which basically tells me the sequence diverges to $\infty$

Answer 1

It does diverges because we have $$\left(-\frac{\sqrt{7}}{3}\right)^{-n}=\left[\left(-\frac{\sqrt{7}}{3}\right)^{-1}\right]^{n}=\left(-\frac{3}{\sqrt{7}}\right)^{n}$$ Which, because $3^2=9$ and $2^2=4$ we have $2<\sqrt{7}<3$, the important fact being $\sqrt{7}<3$ making $\left|-3/\sqrt{7}\right|>1$. Thus it diverges.

Side Note I wouldn't recommend resorting to things like the ratio check to see if a expression diverges as $n\to\infty$ because things like the ratio test are used to check if infinite series converge, and although an infinite series converging implies is summand going to zero, a summand can still go to zero without the series converging.

Answer 2

It's to the power $-n$, so you should consider reciprocal. Note that for $a^n$ to converge to $0$, we must have $|a| < 1$, not just $a<1$.