Krdytkyu

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.

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{1pm (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
$d=1$.

First consider that $a_n$ may have a limit $L$. If $L$ exists then $L=lim_{nto infty}a_{n+1}=lim_{nto 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 $a_{n+1}>a_niff a_{n+1}^2>a_n^2iff 2+a_n>a_n^2,$ 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 $2+a_n>a_n+a_n=(2)(a_n)>(a_n)(a_n)=a_n^2.$