# relations between the root test and the ratio test

relations between the root test and the ratio test

I know the theorem is correct if they are exist $$\lim\inf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} \leq \lim\inf\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \lim\sup\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$$



Here is the 1st question.

If $$\lim\inf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$$ and $$\lim\sup\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$$ are $\infty$

then, $$\lim\inf\limits_{n\rightarrow \infty} (A_n)^{1/n}$$ and $$\lim\sup\limits_{n\rightarrow \infty} (A_n)^{1/n}$$ are $\infty$?





And 2nd question is $$\lim_{n\rightarrow \infty} \frac{|A_{n+1}|}{|A_n|} = \infty$$ then $\lim_{n\rightarrow \infty} (A_n)^{1/n} = \infty$





Actually, 2nd question looks like easy, but I can't prove yet.

Thanks

Both tests are for series with positive terms; or you should put absolute values around $A_n$ and $A_{n+1}$ everywhere. Otherwise you'll have a problem taking the roots. And division by $0$ is generally frowned upon.
Assuming all the terms are positive, it is true that $$\liminf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n} \leq \liminf\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \limsup\limits_{n\rightarrow \infty} (A_n)^{1/n} \leq \limsup_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$$ in the extended sense: the limits are allowed to take value $+\infty$. (By the way, this is common with $\liminf$ and $\limsup$ anyway.
The proof is not really different from the finite case. Here's a proof of the first inequality. Let $b$ be a number such that $b< \liminf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$. Then we have $A_{n+1}>bA_n$ for all sufficiently large $n$. Applying this iteratively, you'll get a lower bound of the form $A_n\ge C b^n$ for all sufficiently large $n$. This implies $\liminf\limits_{n\rightarrow \infty} (A_n)^{1/n} \ge b$. Since $b$ was an arbitrary number such that $b< \liminf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$, the conclusion $\liminf\limits_{n\rightarrow \infty} (A_n)^{1/n}\ge \liminf\limits_{n\rightarrow \infty} \frac{A_{n+1}}{A_n}$ follows.
• @user143993 Suppose the contrary: $A>B$. Take $b$ between $A$ and $B$. Then $b<A$, but $b>B$. – user147263 Jun 8 '14 at 6:17