0

A known theorem states that for a sequence of positive real numbers $\{x_n\}$: $$\lim_{n \to \infty} \frac{x_{n+1}}{x_n} = L \implies \lim_{n \to \infty} \sqrt[n]{x_n} = L.$$

I suspect that the reverse implication is false, but have not yet produced a counterexample.

My suspicion is based on the fact that the convergence of the sequence of $n$-th roots only gives $(L+\varepsilon)^n > x_n > (L - \varepsilon)^n$ for every $\varepsilon$ and sufficiently large $n$, which in turn only implies the following $$\left( 1 + \frac{2 \varepsilon}{L-\varepsilon} \right)^n = \left( \frac{L + \varepsilon}{L - \varepsilon} \right)^n > \frac{x_{n+1}}{x_n} > \left( \frac{L- \varepsilon}{L + \varepsilon}\right)^n = \left( 1 - \frac{2 \varepsilon}{L+\varepsilon} \right)^n,$$ which bounds the sequence of the ratios between two divergent sequences, no matter the size of $\varepsilon$.

Hopefully I have not missed some extremely obvious counterexample, but I have an eerie feeling about that ...

EDIT: Ach yes, it seems like my last remark was accurate.

Trebsch
  • 16

1 Answers1

2

Take $x_n=1$ for $n$ odd and $x_n=2$ for $n$ even. Then the $n-th$ root tends to $1$ but the ratio oscillates between $2$ and $\frac 1 2 $.

Kavi Rama Murthy
  • 359,332