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

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 $$X(n)=\begin{cases}\sqrt{1+2X(n-1)}&n\neq1\\1&n=1\end{cases}$$

• Did you mean $X_{n+1}=\sqrt{1+2X_n}$? Dec 2 '19 at 7:41
• $\sqrt {1+(2)(4)} =\sqrt 9 =3 <4$. Dec 2 '19 at 7:42
• Yes I meant this Xn+1=√1+2Xn..a typing error .so how do I do it Dec 2 '19 at 7:49
• Possible duplicate of Show that $\{a_n\}$ is an increasing sequence and bounded from above Dec 3 '19 at 11:39

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 $$n = k+1$$.

• Okay but how does the proof work using mathematical induction Dec 2 '19 at 8:05
• See my latest edit Dec 2 '19 at 8:18