Double neitz/dds ujelly

This isn't technically right, but I know where it converges.
There is probably a way to "show" it converges, I'm just assuming it does.

Q11a)

a(n) = sqrt (3 + a(n-1))
a(n+1) = sqrt( 3 + sqrt (a(n))

a(n+1) = sqrt (3 + sqrt (3 + a(n-1)))

a(n+1) - a(n) = sqrt (3 + sqrt(3 + a(n-1))) - sqrt (3 + a(n-1))

If we assume series converges, a(n+1) - a(n) = 0

sqrt(3 + a(n-1)) = sqrt(3 + sqrt(3 + a(n-1)))

Square both sides

3 + a(n-1) = 3 + sqrt (3 + a(n-1)

a(n-1) = sqrt (3 + a(n-1))

Square again

a(n-1)^2 = 3 + a(n-1)

a(n-1)^2 - a(n-1) -3 = 0

a(n-1) = 2.302775637

Thats where your series converges.

What I've done is incorrect though, I find a(n-1) instead of a(n).
Unless we assume a(n-1) = a(n). But then I just assumed convergence instead of proving it.
I know it involves subtracting something like a(n+1) away from a(n) and showing it equals something.

On a side note, nobody uses this stuff in the real world anyway.

I HAVE NICE IDEA.