# Superior limit of n-th roots is less than sup lim of fractions.

Let's assume the next:

$$x_k \geq 0$$ $$L_1 = \lim_{n\rightarrow \infty} \sup_{k > n} {x_k}^{\frac{1}{k}}$$ $$L_2 = \lim_{n\rightarrow \infty} \sup_{k > n} \frac{x_{k+1}}{x_k}$$

I need to prove: $$L_1 \leq L_2$$

My ideas:

1) It looks like test for convergence. (if $L_1 \le 1$ or if we test something like {$x_k / (2 * L_1^k)$} )

2) We may try to build counterexample with propety $\forall n \in Nat, {x_k}^{\frac{1}{k}} > \frac{x_{k+1}}{x_k}$. It follows that $x_k \le x_2^{k-1}$. Then there is 3 cases $x_2 (>/< / =)1$. What must be next?

• Welcome to Stack Exchange Math! – bluemaster Sep 17 '17 at 20:17

If $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}=L$, then $\lim_{n\to\infty} (x_n)^{1/n}=L$