(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.
(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.
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.
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 $\displaystyle x^2$ ?