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Here is a solution to a limit of a sequence, but i feel like the squeeze theorem is really needed to get the right answer, but i cant see how excactly it should be used.Thank you for any comment $!$. \begin{align} \delta_{n} & = \frac{1}{2n}\,\sqrt[n]{\,{1^{n} + 2^{n} + \cdots + \left(2n\right)^{n}}\,}\, \\[2mm] & = \,\sqrt[n]{\,{\left(\frac{1}{2n}\right)^{n} + \left(\frac{2}{2n}\right)^{n} + \cdots + \left(\frac{2n}{2n}\right)^{n}}\,}\,\,\,\to 1 \end{align}

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By using $k\leq 2n$ we have

$$\frac{1}{2n}\sqrt[n]{(2n)^n}\leq\frac{1}{2n}\sqrt[n]{1^n+\dots+(2n)^n}\leq\frac{1}{2n}\sqrt[n]{(2n)^n+\dots +(2n)^n}$$

from where we get $$ 1\leq S_n\leq\sqrt[n]{2n}.$$ Since $\lim_{n\to\infty} \sqrt[n]{2n}=1$, we have $$\lim_{n\to\infty} S_n=1.$$

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  • $\begingroup$ This makes perfect sense actually.Big thanks! $\endgroup$
    – GGG
    Commented Jan 16, 2021 at 18:15

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