Show that a sequence is bounded above

[Qwertyuiop[]]

Hi, a sequence is defined by $u_0=0$ and for positive values of n, $u_{n+1}=\sqrt{3u_n+4}$. 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: $u_n\:\le 4$.

Base: $u_0\:=0\:\le 4$ so true.

Assume $u_n\le 4\:for\:some\:n\:\: \in N$, we have to show $u_{n+1}\le 4\:is\:true.$
We are suposing $u_n\le 4$ is true so i did: $u_n\le 4\\ \frac{\left(u_{n+1}\right)^2-4}{3}\le 4\\ \left(u_{n+1}\right)^2\le 16\\ u_{n+1}\le 4$.
I used the $u_{n+1}$ equation and made $u_n$ 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 $x < 4$ then $\sqrt{3x+4} < 4$.

 

[BigBeachBanana]

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

 

[Qwertyuiop[]]

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

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 ($\sqrt{3x+4}$).

 

[BigBeachBanana]

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

 

[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. $u_{n+1} = \mathbf{-}\sqrt{3u_n+4}$. Personally, I'd assume that unless specifically mentioned the positive branch is implied.