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Given $\alpha>0$. Prove that there exists $N \in \mathbb{N}$ such that $e^{n^{\alpha}}>n$, for $n \in \mathbb{N}$, $n \geq N$.

math123
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  • I have tried using ratio test on $a_n=\frac{e^{n^a}}{n}$ to prove that $\lim(a_n)=\infty$, but it seems that my method does not work. Or maybe someone can prove it? – math123 May 02 '13 at 02:52

1 Answers1

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Choose $k$ such that $k\alpha>1$. Then $$e^{n^\alpha}=\frac{(n^{\alpha})^0}{0!}+\cdots+\frac{(n^{\alpha})^k}{k!}+\cdots \ge \frac{1}{k!}n^{\alpha k}$$

Since $\alpha k>1$, for $n$ sufficiently large $\frac{1}{k!}n^{\alpha k}>n$.

vadim123
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