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If $x_n<\infty, \forall \enspace n\in \mathbb{N}$ and $\lim_{n\to\infty}\sqrt[n]{x_n}=l, \enspace l>0$ , show that $\lim_{n\to\infty}\sqrt[n]{(n+1)x_{n+1}}=l$

Following is my approach:
$$\lim_{n\to\infty}\sqrt[n]{(n+1)x_{n+1}}=\left(\lim_{n\to\infty}(n+1)^{1/n}\right).\left(\lim_{n\to\infty}x_{n+1}^{1/n}\right)$$

Now $\lim_{n\to\infty}(1+n)^{1/n}=\lim_{n\to\infty}(n^{1/n})\left(1+\frac{1}{n}\right)^{1/n}=1$
I am stuck in the following step and unable to proceed further :
$$\lim_{n\to\infty}x_{n+1}^{1/n}$$
I don't have any idea how to evaluate this step.

Hamada Al
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sameed hussain
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1 Answers1

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Let $\epsilon>0$. There exists $n_0$ such that $$l-\epsilon<x_n^{\frac{1}{n}}<l+\epsilon$$ if $n>n_0$.

By @KaviRamaMurthy's hint, for $n>n_0$, $$(l-\epsilon)(l-\epsilon)^\frac{1}{n}<x_{n+1}^\frac{1}{n}<(l+\epsilon)(l+\epsilon)^\frac{1}{n}\mbox{.}$$ Letting $n\rightarrow\infty$, we get $$l-\epsilon\leq\limsup x_{n+1}^\frac{1}{n}\leq1+\epsilon\mbox{.}$$ Since $\epsilon$ is arbitrary, $\limsup x_{n+1}^{1/n}=l$. Similarly, $\liminf x_{n+1}^{1/n}=l$, so that $\lim x_{n+1}^{1/n}=l$.