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I encountered a problem when doing my real analysis assignment. The question is:

Assume $x_n>0$ for every $n\in\mathbb{N}$. Show that $\underset{n\rightarrow\infty}{\lim}{\sup}\sqrt[n]{x_n}\leq \underset{n\rightarrow\infty}\lim{\sup}\frac{x_{n+1}}{x_n}.$

I have no ideas how to prove it. Can someone help or give a hint about this?

user26857
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user541649
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  • I'd take logarithms. – Angina Seng Oct 06 '19 at 10:57
  • There are you have 2 formulas to calculate radius of convergence. Root formula - Cachee's Formula; Fraction formula - d'Alambier's Formula. But this formulas actually without the "sup" function – Egor Oct 06 '19 at 11:01
  • And that's strange. You haven't any expression under 'lim'. n->oo should be here. Under 'sup' you can write something about 'x' – Egor Oct 06 '19 at 11:02
  • Another one: https://math.stackexchange.com/questions/69386/inequality-involving-limsup-and-liminf-liminfa-n1-a-n-le-liminf. – Martin R Oct 06 '19 at 11:03
  • Thank you for your comments here, I 'll check it. – user541649 Oct 06 '19 at 11:07

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