Limit infina of ratio test and root test

For a strictly positive sequence, prove that $\displaystyle \liminf (a_j^{\frac{1}{j}}) \geq \liminf (\frac{a_{j+1}}{a_j})$.

I get that it's trying to say that the root test converges faster than the ratio test (hence the greater limit inf), but beyond that, I am drawing a blank. I don't even know where to begin.

Re: Limit infina of ratio test and root test

Hint: For $\displaystyle x>0$, $\displaystyle \lim_{n\to \infty} \sqrt[n]{x}=1$. In particular, $\displaystyle \liminf_{n\to \infty} \sqrt[n]{x}=\lim_{n\to \infty} \sqrt[n]{x}=1$.

Now think about why $\displaystyle \liminf \left(\frac{a_{n+1}}{a_n}\right)\leq 1$. (It may be useful to also think about why $\displaystyle \limsup \left(\frac{a_{n+1}}{a_n}\right) \geq 1$). In particular, what happens if the sequence is eventually constant? What if it never converges?

As an aside, this:

Quote:

I get that it's trying to say that the root test converges faster than the ratio test (hence the greater limit inf)

is not the way to think about it. I'm not even sure if makes sense. Anyways, the following is true for $\displaystyle a_n >0$:

$\displaystyle \liminf \left(\frac{a_{n+1}}{a_n}\right)\quad\leq\quad \liminf \sqrt[n]{a_n}\quad\leq \quad \limsup \sqrt[n]{a_n} \quad \leq\quad \limsup\left(\frac{a_{n+1}}{a_n}\right)$

Re: Limit infina of ratio test and root test

I still do not understand. At all.

Re: Limit infina of ratio test and root test

Do you understand the definition of liminf? Can you see that if a sequence a_n converges, liminf a_n = lim a_n = limsup a_n?

Re: Limit infina of ratio test and root test

Infinum means greatest lower bound. I.e., if A = (0,1), inf A = 0, because 0 <= all elements in A, and 0 is the greatest possible number that satisfies that inequality.

Lim inf x_n = lim (n -> infinity) {a_n, a_n+1, a_n+2, ...}. Basically, lim inf is the infinum of the sequence infinitely far from the starting element.

All of that I get. What I don't get is how to put it together in a meaningful way on this proof.

Re: Limit infina of ratio test and root test

If the sequence$\displaystyle a_n$ is not eventually constant, there are infinitely many points where $\displaystyle a_{n+1} < a_n$. Then $\displaystyle \limsup \frac{a_{n+1}}{a_n}\leq 1 = \liminf \sqrt[n]a_n = \lim \sqrt[n]a_n$. The case where $\displaystyle a_n$ is eventually constant is obvious.