# Compute: $\lim_{n\to\infty} \{ (\sqrt2+1)^{2n} \}$

Compute the following limit:

$$\lim_{n\to\infty} \{ (\sqrt2+1)^{2n} \}$$ where $\{x\}$ is the fractional part of $x$. I need some hints here. Thanks.

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I think $\{(\sqrt2+1)^{2n}\}=2^n\sqrt2$ – Babak S. Jun 22 '12 at 11:42

Consider $$(\sqrt2+1)^{2n} + (\sqrt2-1)^{2n}$$
Try to show that it is an integer and hence this fractional part you are looking for is $1 - (\sqrt2-1)^{2n}$ Now the limit becomes easy.