Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I think $\{(\sqrt2+1)^{2n}\}=2^n\sqrt2$ – Babak S. Jun 22 '12 at 11:42
up vote 18 down vote accepted

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.

share|cite|improve this answer
got it! Thanks! – user 1618033 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.