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If $(x_n)_{n\ge0}$ is positive and increasing, find $\lim\limits_{n\rightarrow\infty} \sqrt[n]{x_1^n+...+x_n^n}$.

Any solution or help is welcome! Thanks in advance!

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    For the record, this is the fourth question with zero context you post today. My suggestion: stop the flow and start adding some personal input. – 2017-02-03
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    @Did Stalker much? Anyway. I'll jeep that in mind, thanks. – 2017-02-03
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    @theSongbird Nah, many high rep users simply remember particular users (usually through experience with them) – 2017-02-03

1 Answers 1

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Let us call $L=\lim_{n\to +\infty} x_n$ ($L$ could be $+\infty$) and $a_n=\sqrt[n]{x_1^n+\cdots+x_n^n}$.

We have, because $x_n>0$ for all $n$ $$x_n=\sqrt[n]{x_n^n}\leq \sqrt[n]{x_1^n+\cdots+x_{n-1}^n+x_n^n}=a_n$$

and, as $x_n$ is increasing: $a_n=\sqrt[n]{x_1^n+\cdots+x_n^n}\leq \sqrt[n]{x_n^n+\cdots+x_{n}^n}=\sqrt[n]{nx_n^n}=\sqrt[n]{n} \cdot x_n$

So we get $$x_n\leq a_n \leq \sqrt[n]{n}\cdot x_n$$

Now note that $\lim_{n\to\infty} \sqrt[n]{n}=1$, so $a_n$ is between two sequences whose limit is $L$. Using the squeeze theorem: $$\lim_{n\to +\infty} a_n=L=\lim_{n\to +\infty} x_n$$