1
$\begingroup$

I'm trying to solve this problem from the text but I can't seem to figure out how to exactly approach it. It says,

Find the limit of the convergent sequence, $(\sqrt{2}-1)^n$ as n goes from 1 to $\infty$

I'm assuming I have to find the value of the sequence as n goes to infinity by applying the limit, but since its to the power n, I'm having bit of a trouble simplifying it. Please correct me if I'm wrong.

$\endgroup$
  • 2
    $\begingroup$ Hint: Since $1\lt\sqrt2\lt2$, $0\lt\sqrt2-1\lt1$ $\endgroup$ – robjohn Jan 15 '14 at 5:36
3
$\begingroup$

Note $0<\sqrt 2-1<1$. What do you know about $r^n$ if $|r|<1$?

$\endgroup$
2
$\begingroup$

One unique hint:

$$\lim_{n\to\infty} x^n=0\;\;\;\forall\,|x|<1$$

$\endgroup$
  • $\begingroup$ ...to rule them all and in the Darkness, bind them? (+1) $\endgroup$ – Did Jan 15 '14 at 7:40
  • $\begingroup$ "...in the Land of Mordor where the Shadows lie" . And even without a ring, in this case...:) $\endgroup$ – DonAntonio Jan 15 '14 at 10:40
0
$\begingroup$

One more hint:

Each number is positive and each number is less than $1/2$ the previous number, since $1 < \sqrt{2} < 1.5$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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