Let $x_1 = 1$ and $x_{n+1}$ = $\sqrt{{x_n}^2 + \frac{1}{(x_n)^2}}$

Prove by mathematical induction that for all $n ≥ 1, 1 ≤ x_n ≤ \sqrt n$

I tested $P(1)$ and found $1 \leq 1 \leq 1$, which holds up.

Now I need to show $P(n+1)$.

By the inductive assumption, $ 1 \leq x_k \leq \sqrt k$ where $k=n$ and $k \geq 1$.

Then I did this, but I'm not sure what to do from here.

$1 \leq x^2_k \leq k $

Suggestions would be helpful.

  • $\begingroup$ Can you find upper and lower bounds for ${x_{k+1}^2}$? $\endgroup$ – player3236 Oct 12 '20 at 1:18
  • $\begingroup$ @player3236 The lower bound would be 2 where k = 1, and the upper bound would be infinity. That means the lowest possible value for $x_k+1$ would be $\sqrt 2$. $\endgroup$ – mathstudent288 Oct 12 '20 at 1:25
  • $\begingroup$ I mean for some specific $k$, using $1 \le x_k^2 \le k$. $\endgroup$ – player3236 Oct 12 '20 at 1:26

The induction hypothesis is $1\le x_n\le\sqrt n$. Then $1\le x_n^2\le n$, so $0<\frac{1}{x_n^2}\le 1$. Therefore $1\le x_n^2+\frac{1}{x_n^2}\le n+1$, and finally $1\le\sqrt{x_n^2+\frac{1}{x_n^2}}\le \sqrt{n+1}$. So $1\le x_{n+1}\le\sqrt{n+1}$.

  • $\begingroup$ Thank you so much! This is understandable but only intuitive for me after I have seen the solution. I wasn't expecting it to be so simple. $\endgroup$ – mathstudent288 Oct 12 '20 at 1:43

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.