Induction of infinite series

Question

A bit confused by this problem.

Let {an} be the sequence defined recursively by

a0 = 0 an+1 = square root of (2+an)

a) Show that if an < 2 , then an+1 < 2. Conclude by induction that an<2 for all n.

I'm pretty sure I got this part, I just showed that square root (2+2) was equal to 2 and that the square root (2+an) would never be equal to or higher than 2 because an is less than 2.

I was unable to get part b and c, however, which ask:

b) Show that if an < 2, then an < an+1 . Conclude by induction that an is increasing.

c) Use (a) and (b) to conclude that L = Lim as n goes to infinity of an exists. Then compute L by showing that L=Square root (2+L)

Thanks for any help.

Answer 1

For (b), you want to compare $a_n$ and $a_{n+1}$. Suppose that what you're trying to prove fails - this means $a_n<2$ but $\sqrt{a_n+2}\le a_n$. Squaring both sides and moving things around gives $$0\le a_n^2-a_n-2=(a_n-2)(a_n+1).$$ For a product $XY$ to be nonnegative, either both $X$ and $Y$ are nonnegative or both $X$ and $Y$ are nonpositive. But $a_n+1$ is always going to be nonnegative, since it's the square root of something; so for the inequality above to be true, the other term $a_n-2$ also has to be nonnegative.

What does it mean for $a_n-2$ to be nonnegative?

For (c), what do you know about limits already?

Answer 2

Observe that if $a_n < 2$, then: \begin{align*} a_{n+1} &= \sqrt{2 + a_n} &\text{by definition} \\ &> \sqrt{a_n + a_n} &\text{since $2 > a_n$} \\ &= \sqrt{2a_n} \\ &> \sqrt{a_na_n} &\text{since $2 > a_n$} \\ &= a_n &\text{assuming $a_n \geq 0$} \end{align*}