a_1 = a, a_n+1 = SQRT( a + a_n) monotonic? bounded? ...

[dts5044]

Let a[sub:76u9vo7o]1[/sub:76u9vo7o] = a, where a > 0 (constant), and a[sub:76u9vo7o]n+1[/sub:76u9vo7o] = SQRT(a + a[sub:76u9vo7o]n[/sub:76u9vo7o]) for n >=1.

Is the sequence monotonic, bounded, convergent? If yes, find its limit.

Again, I am pretty sure I need to prove it by induction but don't know where to start. Please help.

 

[pka]

Yes this is a monotonic sequence.
BUT depending on the value of a it may be increasing or decreasing.
\(\displaystyle a_2 = \sqrt {3 + 3} < a_1 < 3\)

Suppose that
\(\displaystyle \begin{gathered} a_{k + 1} < a_k < 3 \hfill \\ a_{k + 1} + a < a_k + a < 3 + a \hfill \\ \sqrt {a_{k + 1} + a} < \sqrt {a_k + a} < \sqrt {3 + a} \hfill \\ \therefore \quad a_{k + 2} < a_{k + 1} < 3 \hfill \\ \end{gathered}\)
That proves in inductively.
As in my other answer solve for L: \(\displaystyle L = \sqrt {a + L}\)