Step on a proof that the root test is stronger than the ratio test.

Question

The following is from the third chapter of baby Rudin:

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I can't seem to understand why $\sqrt[n]{c_n}\leq \sqrt[n]{c_N\beta ^{-N}}\beta $ implies $\sqrt[n]{c_n}\leq \beta $. This seems to presuppose that $\sqrt[n]{c_N\beta ^{-N}}\leq 1$, or that $\beta^N\geq c_N$, but I can't see why that's the case.

Answer 1

$\sqrt[n]{c_n}\leq \sqrt[n]{c_N\beta ^{-N}}\beta$ implies that $\limsup_{n\to\infty}\sqrt[n]{c_n}\leq \limsup_{n\to\infty}\sqrt[n]{c_N\beta ^{-N}}\beta=\beta$, since $\lim_{n\to\infty}\sqrt[n]{a}=1\quad\forall a>0$

Answer 2

It does not imply that $(c_n)^{1/n} \leq \beta$. But what is being used here is the fact that $x^{1/n} \to 1$ as $n \to \infty$ for any $x>0$. Note that the previous inequality holds for $n \geq N$. So fix $N$ and let $n \to \infty$ to see that $(c_N\beta^{-N})^{1/n} \to 1$ ad $ n \to \infty$.