a_1 >= (a + SQRT (a^2 + 4b))/2, a_n+1 = SQRT(a*a_n +b) ...

[dts5044]

Let a, b be two constant real numbers such that a > 0 and a[sup:11d97x91]2[/sup:11d97x91] + 4b > 0

Let a[sub:11d97x91]1[/sub:11d97x91] >= (a + SQRT(a[sup:11d97x91]2[/sup:11d97x91] + 4b))/2 be a given real number, and a[sub:11d97x91]n+1[/sub:11d97x91] = SQRT(a * a[sub:11d97x91]n[/sub:11d97x91] + b) for n >=1

is the sequence a[sub:11d97x91]n[/sub:11d97x91] monotonic? bounded? convergent? If yes, find its limit

I assume you need to use induction, but can't seem to find a starting point. Can someone help?

 

[Deleted member 4993]

Let a, b be two constant real numbers such that a > 0 and a[sup:1tawk0fa]2[/sup:1tawk0fa] + 4b > 0

Let a[sub:1tawk0fa]1[/sub:1tawk0fa] >= (a + SQRT(a[sup:1tawk0fa]2[/sup:1tawk0fa] + 4b))/2 be a given real number, and a[sub:1tawk0fa]n+1[/sub:1tawk0fa] = SQRT(a * a[sub:1tawk0fa]n[/sub:1tawk0fa] + b) for n >=1

is the sequence a[sub:1tawk0fa]n[/sub:1tawk0fa] monotonic? bounded? convergent? If yes, find its limit

I assume you need to use induction, but can't seem to find a starting point. Can someone help?

what are the definitions of:

monotonic? bounded? convergent?

Why/How do you think "induction" will prove any of those?

By the way, pka answered your previous post - and that should provide a strong hint for solving this one. You need to show some of your work here.