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If each $a_n > 0$ for a sequence ${a_n}$, prove that

$$\liminf_{n\to\infty}\left(\frac{a_{n+1}}{a_n}\right)\leq \liminf_{n\to\infty}\sqrt[n]{a_n}\leq \limsup_{n\to\infty}\sqrt[n]{a_n}\leq \limsup_{n\to\infty}\left(\frac{a_{n+1}}{a_n}\right)$$

The center inequality is given by definition of $\liminf_{n\to\infty}$ and $\limsup_{n\to\infty}$

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This is explained in Rudin's Principles of Mathematical Analysis. Check it out. –  Weltschmerz Apr 5 '11 at 21:55
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Please replace $a_n+1$ by $a_{n+1}$. Twice. –  Did Apr 5 '11 at 22:04
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up vote 1 down vote accepted

For the right hand side... suppose $\limsup_{n \rightarrow \infty}{a_{n+1} \over a_n} = c < \infty$. Then if $d > c$, there exists some $N$ such that if $n > N$ you have ${a_{n+1} \over a_n} < d$. Hence multiplying from $n = N$ to $n = M-1$ for some $M > N$ you have ${a_M \over a_N} < d^{M - N}$ or just $a_M < a_N d^{M - N}$. Hence $a_M^{1 \over M} < a_N^{1 \over M}d^{1 - {N \over M}}$. Now take limsups as $M$ goes to infinity.

A similar thing works for the left hand side.

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