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Question Number 128626 by john_santu last updated on 09/Jan/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left\{{(2+\sqrt{2})}^{\mathrm{n}}\right\}=?$$$$\left\{\right\}=\mathrm{fractional}$$$$$$

Commented by liberty last updated on 09/Jan/21

$$\left\{x\right\}=x-\lfloor x\rfloor$$

Answered by mnjuly1970 last updated on 09/Jan/21

$$solution:$$$$t_n:={(2+\sqrt{2})}^n+{(2-\sqrt{2})}^n\stackrel{why?}{\in }\mathbb{N}$$$$because:t_n=\underset{k=0}{\sum ^n}\{\left(\begin{array}{c}n\\ k\end{array}\right){\left(2\right)}^{n-k}{\left(\sqrt{2}\right)}^k(1+{(-1)}^k)$$$$=\underset{k\underset{k\in {\mathbb{N}}_E}{=}0}{\sum ^n}\left\{\left(\begin{array}{c}n\\ k\end{array}\right)2^{n-k+\frac{k}{2}+1}\right\}$$$$=\sum _{k\underset{k\in {\mathbb{N}}_E}{=}0}\left\{\left(\begin{array}{c}n\\ k\end{array}\right)2^{[n+\frac{k}{2}+1\in \mathbb{N}]}\right\}\in \mathbb{N}$$$$butweknow:0<{(2-\sqrt{2})}^n<1$$$$\therefore 0<1-{(2-\sqrt{2})}^n<1$$$${lim}_{n\to \mathrm{\infty }}(1-{(2-\sqrt{2})}^n)=1$$$$\left\{{(2+\sqrt{2})}^n\right\}=\{t_n-{(2-\sqrt{2})}^n\}$$$$=\{t_n-1+1-{(2-\sqrt{2})}^n\}$$$$\therefore {lim}_{n\to \mathrm{\infty }}\left\{{(2+\sqrt{2})}^n\right\}=1$$$$note:\{t_n-1\}=0$$

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