Is ratio test valid for infinite limit?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $n\to \infty.$