Consider the sequence of real numbers defined by the relations: $X(1)=1$ and $X(n+1)=\sqrt{1+2X(n)}$ for $n\geq1$

Question

How can I use the principle of mathematical induction to show that $X(n)<4$ for all $n\geq1$ where $X(n)$ is defined as $$X(n)=\begin{cases}\sqrt{1+2X(n-1)}&n\neq1\\1&n=1\end{cases}$$

Answer 1

Hint: It suffices to note that the range of the function $f: [1,4) \to \Bbb R$ defined by $f(x) = \sqrt{1 + 2x}$ is $[1,3)$.

More specifically: we are showing that $1 \leq X(n) < 4$ for all integers $n\geq 1$, using the fact that $X(n+1) = f(X(n))$ for all $n \geq 1$. The base case is that this holds for $n = 1$. The inductive step is to show that if this holds for $n = k$, then it also holds for $n = k+1$.