Is ratio test valid for infinite limit? If $$\lim_{n\rightarrow\infty} \frac{u_{n+1}}{u_n} = \infty$$ then by ratio test of convergence of series, can I conclude anything?
If $$\lim_{n\rightarrow\infty} \frac{u_{n+1}}{u_n}=\infty$$, then $$\lim_{n\rightarrow\infty} \frac{u_{n+1}}{u_n}>k$$ for any $k>0$.
Then by comparison test, can I say $$\sum_{n=1}^\infty u_n$$ is divergent?
 A: I assume all $u_n>0.$ Because the ratios go to infinity, there exists $N$ such that $\dfrac{u_{n+1}}{u_n} > 2$ for $n\ge N.$ As in the proof of the "finite" ratio test, this implies
$$u_{N+k} > 2^k u_N,\,\, k=1,2,\dots$$
Thus $u_{N+k} \to \infty$ as $k\to \infty.$ This of course implies $u_n \to \infty$ as $n\to \infty.$
A: It must be $+\infty $
if the sequence has a constant sign and
$$\lim_{n\to+\infty}\frac {u_n}{u_{n+1}}=+\infty $$
then
$$\lim_{n\to+\infty}\frac {u_{n+1}}{u_n}=0 <1$$
thus by D'Alembert test, the series $\sum u_n $ converges.
A: The root test says that if
$$
L:=\lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|
$$
exists, then the series $\sum u_n$ is absolutely convergent if $L<1$ and divergent if $L>1$.
Now if $(u_n)_n$ satisfies
$$
\lim_{n\to\infty}\frac{u_n}{u_{n+1}} = \infty,
$$
then
$$
\lim_{n\to\infty}\frac{u_{n+1}}{u_n} = 0
$$
and consequently
$$
\lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right| = 0.
$$
Therefore, by the ratio test, we conclude that $\sum u_n$ is absolutely convergent.
There is not any comparison test at work here.
