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Suppose that $x_1, \ldots, x_n > 0$. How to show that $\lim_{n \to \infty} \sqrt[n]{x_n}=1$ if $\lim_{n \to \infty} x_n = a>0$?

My solution: suppose that $\sqrt[n]{x_n}=1 + \delta_n$. Then $(1+\delta_n)^n = 1 + n \delta_n + \frac{n(n-1)}{2}\delta_n^2 + \cdots + \delta_n^n = x_n$. But we don't know $\delta_n>0$ or not.

Thank you very much.

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  • $\begingroup$ Does it matter if $\delta_n<0$? With the approach you're taking it looks like all that matters is that $\delta_n \to 0$ and $\delta_n\geq-1$ $\endgroup$ – graydad Oct 26 '15 at 3:27
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Since $a=\lim a_n$ exists and is positive, we know there are positive $b,c$ with $0<b<a<c$, such that $b<a_n<c$ for all but finitely many $n$. Hence $b^{1/n}<a_n^{1/n}<c^{1/n}$ for all but finitely many $n$.It suffices to have $\lim b^{1/n}=\lim c^{1/n}=1$ as $n\to \infty$....We can prove that $$\forall r\in R^+ (1=\lim_{n\to \infty}r^{1/n})$$ as follows: Case(1). For $r\geq 1$ we have $r=1+(r-1)\leq (1+(r-1)/n)^n$ for positive integer $n$, by the Binomial Theorem. So $$1\leq r^{1/n}\leq 1+(r-1)/n.$$ Case(2), For $0<r<1$, let $r=s^{-1}.$ We have $s>1$. By case (1) we have $\lim_{n\to \infty}s^{1/n}=1$. So $$1=\lim_{n\to \infty}s^{-1/n}=\lim_{n\to \infty}r^{1/n}.$$

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