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Let $a_n$ be a sequence of real positive numbers. Show $$\liminf_{n\to  \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n\to \infty}\,(a_n)^{1/n}  \leq \limsup_{n\to \infty} \, (a_n)^{1/n} \leq \limsup_{n\to \infty}  \frac{a_{n+1}}{a_n}.$$

Here's the proof I had in mind, and I'm interested to know if anyone knows a simpler one, or if anyone can find a gap in my logic.

Suppose $\liminf_{n\to \infty}(a_n)^{1/n}=L_1$. Then we have a subsequence $a_{n_k}$ such that $(a_{n_k})^{1/n_k}\to L_1$. Now 

$$(a_n)^{1/n}=\left( \frac{a_n}{a_{n-1}} \frac{a_{n-1}}{a_{n-2}} \dotsb \frac{a_2}{a_1} a_1 \right)^{1/n}\\ \log\left( \sqrt[n]{a_n} \right)=\frac{\log\left(\frac{a_n}{a_{n-1}}\right) +\dotsb +  \log\left(\frac{a_2}{a_1}\right) + \log(a_1)}{n}.$$

If $\log(\sqrt[n_k]{a_{n_k})}\to \log(L_1)$, then we can show by contradiction that $\forall \epsilon$ there are arbitrarily high $n_l$ such that $$\frac{a_{n_l}}{a_{n_l -1}} < L_1 + \epsilon $$

If not, then eventually the averages on the right-hand side would be too large.

Martin Sleziak
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Eric Auld
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1 Answers1

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Looks good. As you see from the last indirect part, it might be better to start from the ratio test with $L_2=\liminf_{n\to\infty} \left|\tfrac{a_{n+1}}{a_n}\right|$. Then use that for any $ϵ>0$ and almost all $n$, $$\left|\tfrac{a_{n+1}}{a_n}\right|\ge L_2-ϵ$$ and construct from that a lower boundary for the root test.

Lutz Lehmann
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