Conclusions from the ratio test for sequences?

Question

For the test I know if $\frac{x_{n+1}}{x_n}$ converges to $L$ and $L<1$, then xn converges to 0. Does this still apply if L is negative, and what about if $L\ge1$? Is it the same for series where if $L=1$ the test is ambiguous, and if $L>1$, the sequence will diverge?

Answer 1

Take your $$L=\frac{x_{n+1}}{x_n}$$

For $x_0=k$, with $L$ tending to its final value, your sequence becomes roughly $$k,Lk,L^2k,L^3k...$$

If $L=1$, this means that $$x_{n+1}=x_n$$ Hence your sequence becomes $k,k,k,k,k,k,k,k....$

If $L<1$, note that $L^{n+1}<L^n$ for $n\ge1,n\in \Bbb Z$

Hence via our general form, $\lim_{n\to\infty}{(L^nk)}=0$

For $L>1$, our inequality reverses. We get $L^{n+1}>L^n$ for $n\ge1,n\in \Bbb Z$

And so $\lim_{n\to\infty}{(L^nk)}=\infty$ and this diverges

Answer 2

The ratio test is about the limit$$L=\lim_{n\to\infty}\left|\frac{x_{n+1}}{x_n}\right|.$$Since each $\left|\frac{x_{n+1}}{x_n}\right|$ is non-negative, $L\geqslant0$. Yes, there is nothing you can deduce if $L=1$ and the sequence diverges if $L>1$.