Show that a sequence is bounded above

[Qwertyuiop[]]

Hi, a sequence is defined by [imath]u_0=0[/imath] and for positive values of n, [imath]u_{n+1}=\sqrt{3u_n+4}[/imath]. Show that the sequence is bounded above 4.
I think i got the answer but i'm not sure if the working is correct. I used induction to get the answer but there is one part in the process i am not sure if it's correct.

so the sequence is bounded above 4: [imath]u_n\:\le 4[/imath].

Base: [imath]u_0\:=0\:\le 4[/imath] so true.

Assume [imath]u_n\le 4\:for\:some\:n\:\: \in N[/imath], we have to show [imath]u_{n+1}\le 4\:is\:true.[/imath]
We are suposing [imath]u_n\le 4[/imath] is true so i did: [imath]u_n\le 4\\ \frac{\left(u_{n+1}\right)^2-4}{3}\le 4\\ \left(u_{n+1}\right)^2\le 16\\ u_{n+1}\le 4[/imath].
I used the [imath]u_{n+1}[/imath] equation and made [imath]u_n[/imath] the susbject, that's the part I think is probably wrong, are you allowed to do this? And is my working correct? If not, how i should do i question like this? thank you.

 

[blamocur]

Looks right to me. But I think there is a more straightforward way to show that if [imath]x < 4[/imath] then [imath]\sqrt{3x+4} < 4[/imath].

 

[BigBeachBanana]

@blamocur is referring to complete (strong) induction. That is for all [imath] k \leq n[/imath], then [imath]u_{n} \leq 4[/imath].
[math]u_{n} \leq \sqrt{3\cdot 4 +4 }=4[/math]

 

[Qwertyuiop[]]

@blamocur is referring to complete (strong) induction. That is for all [imath] k \leq n[/imath], then [imath]u_{n} \leq 4[/imath].
[math]u_{n} \leq \sqrt{3\cdot 4 +4 }=4[/math]

so we just substitute u=4 in the equation and see if satisfies the inequality?

 

[blamocur]

so we just substitute u=4 in the equation and see if satisfies the inequality?

In the general case you would need to show that the function is monotonic, which is kind of obvious in your case ([imath]\sqrt{3x+4}[/imath]).

 

[BigBeachBanana]

Alternatively, assuming the sequence is monotonically increasing and convergent.
[math]\lim u_{n+1} = \lim u_{n} \iff u_{n} = \sqrt{3\cdot u_{n}+4} \iff (u_{n})^2 = 3u_{n}+4 \iff u_{n} =-1, 4[/math]
Since [imath]u_0 = 0[/imath] and the sequence is monotonically increasing and convergent, we omit [imath]-1[/imath].

 

[blamocur]

=0 and the sequence is monotonically increasing and convergent, we omit −1-1−1.

Keeping -1 as a solution would require using negative square root, i.e. [imath]u_{n+1} = \mathbf{-}\sqrt{3u_n+4}[/imath]. Personally, I'd assume that unless specifically mentioned the positive branch is implied.