Question

Let a1 >sqroot 2, and let an+1=an/2+1/an for n=1,2,3..
a) Show that {an} is decreasing and an> sq 2 for n>or equal to 1
b) Why must {an} converge?
c) find lim an as n approaches infinity

Answer 1

is it
\[a_{n+1}=\frac{a_n}{2}+\frac{1}{a_n}\]?

Answer 2

yes and and a1 \[\sqrt{2}\]

Answer 3

\[a_{n+1}=\frac{1}{2}a_n+\frac{1}{a_n},~a_1=\sqrt2\]
To show it's decreasing, try to prove by induction:
\[\begin{align*}
n=1:&\\
&a_2=\frac{1}{2}a_1+\frac{1}{a_1}\\
&a_2=\frac{\sqrt2}{2}+\frac{1}{\sqrt2}\\
&a_2=\frac{2+\sqrt2}{2\sqrt2}<\sqrt2=a_1
\end{align*}\]
So the claim that it's decreasing holds for the base case. Assume it holds for \(n=k\) and show that it holds for \(n=k+1\).

Answer 4

And so it converges because the limit is less than 1?

Answer 5

or equal to 1

Answer 6

Well, no it's supposed to converge because it's monotonically decreasing (what you're showing in part (a)) and it's bounded (less than \(\sqrt2\), also a result of part (a)).

Answer 7

If its bounded by less than \[\sqrt{2}\] then it will keep decreasing?

Answer 8

The sequence is decreasing because you show that \(a_{n+1}<a_n\) for all \(n\) (as a result of the induction proof). You also know that \(a_n=\sqrt2\), so \(a_{n+1}<a_n=\sqrt2 \) for all \(n\). You know that every number in the sequence is positive, so the sequence is bounded between \(0\) and \(\sqrt 2\).

Answer 9

(You still have yet to show that it's decreasing, by the way. I've started the proof for you, but it's only half done.)

Answer 10

Ok that makes sense! Thanks so much!