0

I'm trying to show that $a_n$ is increasing where $a_1=1$ and $a_{n+1}$ = $ \sqrt{a_n+2}$

I proceeded by induction so showed that $a_{n+1}>a_n$ for all n greater than 1 So I showed it was true for n=1

Then assumed $a_{k+1}>a_k$ for some k greater than 1

Then I add two to both sides and square root to show it's true for k+1.

So $a_{k+1}+2>a_k+2$

$ \sqrt{a_{k+1}+2}> \sqrt{a_k+2}$

Which is $a_{k+2}>a_{k+1}$ So is this ok to show this statement?

Also how would you show that a_n has a limit and compute $lim_na_n$

I think to compute the limit you need to prove that if $b_n$ goes to b then $ \sqrt{b_n+2}$ goes to $ \sqrt{b+2}$ thank you for your help.

Johnf
  • 25
  • Your argument is incorrectly structured. You prove it for $n = 1$, then, assuming it is true for $n$, show it is true for $n + 1$. – ncmathsadist Nov 26 '18 at 01:43
  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Nov 26 '18 at 01:43
  • 1
    Please [edit] the question accordingly :) – Shaun Nov 26 '18 at 01:44
  • Ok, I don't understand is that not what I have done – Johnf Nov 26 '18 at 01:47
  • Hi, could you edit your question to make it a little clearer by including some of the steps, thanks. – Carlos Bacca Nov 26 '18 at 02:23

2 Answers2

0

If $a_{n+1} =\sqrt{a_n+2} $ then $a_{n+1}^2 =a_n+2 $ so $a_{n+1}^2-a_n^2 =a_n+2-a_n^2 =-(a_n^2-a_n-2) =-(a_n-2)(a_n+1) $.

Therefore, if $0 < a_n < 2$ then $a_{n+1}^2-a_n^2 \gt 0$ since $a_n-2 < 0$ and $a_n+1 > 0$.

Also, if $0 < a_n < 2$ then $\sqrt{a_n+2} \lt \sqrt{4} = 2$.

For $a_1 = 1$ we have $a_n$ increasing and bounded so $a_n \to L$ for some $1 < L \le 2$.

Therefore $L^2 = L+2$ so $L = 2$ or $L = -1$.

Since $L > 1$, we must have $L = 2$.

More generally, if $a_{n+1} =\sqrt{a_n+k} $ then $a_{n+1}^2 =a_n+k $ so $a_{n+1}^2-a_n^2 =a_n+k-a_n^2 =-(a_n^2-a_n-k) $.

This factors if $1+4k$ is a square, $(2d+1)^2 = 4k+1$ (since $4k+1$ is odd). The roots of $x^2-x-k = 0$ are then $\dfrac{1\pm (2d+1)}{2} =\dfrac{2d+2, -2d}{2} =d+1, -d $.

Therefore $a_{n+1}^2-a_n^2 =-(a_n^2-a_n-k) =-(a_n-d-1)(a_n+d) $.

If $0 < a_n \le d+1$ then $a_{n+1}^2-a_n^2 \gt 0$ since $a_n-d-1 < 0$ and $a_n+1 > 0$.

Also, if $0 < a_n < d+1$ since $(2d+1)^2 = 4k+1$ so that $k = d^2+d$, then $\sqrt{a_n+k} \lt \sqrt{d+1+d^2+d} = \sqrt{d^2+2d+1} = d+1$.

For $0 < a_1 \lt d+1$ we have $a_n$ increasing and bounded so $a_n \to L$ for some $1 < L \le d+1$.

Therefore $L^2 = L+k$ so $L = d+1$ or $L = -d$.

Since $L > 0$, we must have $L = d+1$.

For the original problem, $k=2$ gives $d=1$.

marty cohen
  • 107,799
0

First consider that $a_n$ may have a limit $L$. If $L$ exists then $L=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\sqrt {2+a_n}\,=\sqrt {2+L}. $

So if $a_n$ is strictly increasing with $a_1=1$ and if $L$ exists then $0<L=\sqrt {2+L}\,$ so $L=2.$ So prove that $2>a_{n+1}>a_n >0$ for all $n$ by induction:

(i). We have $2>a_1>0.$

(ii). If $2>a_n>0$ then $4>2+a_n>0$ so $2>\sqrt {2+a_n}\,=a_{n+1}>0$. So $2>a_n>0$ for all $n$ by induction.

(iii). Since $a_{n+1}>0$ by (ii), we have $a_{n+1}>a_n\iff a_{n+1}^2>a_n^2\iff 2+a_n>a_n^2,$ and we do have $2+a_n>a_n^2$ when $2>a_n>0.$ So $a_{n+1}>a_n$ for all $n$ by induction.

So $a_n$ is strictly increasing and is bounded above by $2$ so it has a limit $L$. And by the first paragraph, $L=2.$

BTW, in (iii), if $2>a_n>0$ then $2+a_n>a_n+a_n=(2)(a_n)>(a_n)(a_n)=a_n^2.$

DanielWainfleet
  • 57,985