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.

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got it! Thanks! –  Chris's sis the artist Jun 22 '12 at 11:46
It's limit is coming out to be 1, but fractional part can't be 1, so the result is 0 or 1? –  Aang Jun 22 '12 at 11:51
I don't see why the fact that no element in the sequence can be 1 should stop the limit from being 1. –  Wonder Jun 22 '12 at 11:54
okay!, i got your point :).Thanks!! –  Aang Jun 22 '12 at 12:06