1. ## sequence questions

(a) Prove that if 0<a<2, then a<2$\displaystyle \sqrt{a}$<2.

(b) Prove that the sequence $\displaystyle \sqrt{2}$, $\displaystyle \sqrt{2\sqrt{2}}$, $\displaystyle \sqrt{2\sqrt{2\sqrt{2}}}$ converges.

(c) Find the limit of the sequence.

Thanks.

2. Originally Posted by khatz
(a) Prove that if 0<a<2, then a<2$\displaystyle \sqrt{2}$<2.
It isn't true! Must be a typo somewhere.

3. For (b) show that the sequence is increasing (I wont show that here) and bounded.

Clearly $\displaystyle a_1=\sqrt{2}<2$ so assume that $\displaystyle a_n<2$ for some n. Then $\displaystyle a_{n+1}=\sqrt{2a_n}<\sqrt{2*2}=2$ so $\displaystyle a_n<2$ for all n and is thus bounded.

4. Originally Posted by AlephZero
It isn't true! Must be a typo somewhere.
Yes, it should be 2sqrt(a).

5. Originally Posted by khatz
Yes, it should be 2sqrt(a).
Still not true!

Counterexample: $\displaystyle a=1.9 < 2 \implies \sqrt{a}\approx 1.38 \implies 2\sqrt{a} \approx 2.76 > 2.$

6. Originally Posted by khatz
(a) Prove that if 0<a<2, then a<2$\displaystyle \sqrt{a}$<2.

(c) Find the limit of the sequence.

Thanks.
Considering what it leads onto, I think (a) should be

(a) Prove that if $\displaystyle 0<a<2$, then $\displaystyle a< \sqrt{2a}<2$

For (c), if the limit it exists we can say that $\displaystyle \sqrt{2 \sqrt{2\sqrt{2\sqrt{\ldots}}}} = x$

So what can you say about $\displaystyle x^2$ ?