sequence Xn is bounded, monotone, convergent

Question

I need to study if the sequence is bounded, monotone and convergent.

$$X_1=\sqrt 3,\ X_{n+1} = \sqrt{2X_n+3},\ n\in\mathbb{N}$$

Answer 1

It's easy to check by induction that $3$ is an upper bound for $(X_n)$. Clearly this is true for $n=1$. Suppose $X_{n} < 3$. Then $2X_n + 3 < 9$, so $X_{n+1} < \sqrt{9} = 3$.

To show that $(X_n)$ is monotonically increasing, first note that clearly all of the $X_n$ are positive (easy induction argument). Therefore $X_{n+1} > X_n$ if and only if $X_{n+1}^2 > X_n^2$, which you can verify is true if and only if $0 < X_n < 3$.

Since $(X_n)$ is increasing and bounded above, it converges, and you can find the value to which it converges by taking limits of both sides of $X_{n+1} = \sqrt{2X_n + 3}$.

Answer 2

Hint:

• Show by induction that $1 <X_n < 3$ for all $n$,
• Show that the sequence is monotone increasing by calculating $X_{n + 1}^2 - X_n^2 = 2X_n + 3 - X_n^2 = -(X_n - 1)^2 + 4$, use the range of $X_n$ to argue this is greater than $0$.
• Use monotone bounded convergence theorem.