$c_n>0$ in $\mathbb{R}$, prove$$\limsup \frac{c_{n+1}}{c_n} \geqslant \limsup \sqrt[n]{c_n}$$ Tried to present $\sqrt[n]{c_n}=c_1 \frac{c_2}{c_1}...\frac{c_{n+1}}{c_n}$. Seemingly $\sqrt[n]{c_n}$ is some kind of mean of $c_1$ to $c_{n+1}/c_{n}$. However nothing can be done...any hint?
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Let $l := \limsup_{n \to \infty}c_{n+1}/c_{n}$; let $\varepsilon > 0$; then there is some $N \geq 1$ such that $n \geq N$ only if $c_{n+1}/c_{n} < l+\varepsilon$, only if $c_{N+p} < c_{N}(l+\varepsilon)^{p}$ for all $p \geq 1$ by induction, only if $c_{n} < c_{N}(l + \varepsilon)^{n-N}$ for all $n \geq N$, only if $c_{n}^{1/n} < c_{N}^{1/n}(l + \varepsilon)^{1 - \frac{N}{n}}$ for all $n \geq N$, and only if $\limsup_{n \to \infty}c_{n}^{1/n} \leq l + \varepsilon$; this argument holds for every $\varepsilon > 0$, so we have $$ \limsup_{n \to \infty}c_{n}^{1/n} \leq l. $$
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