Chapter 9.5, Problem 48E

Solution

To prove : Using mathematical induction that if $\left\{a_n\right\}$ is the sequence $\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}},\dots$ then $a_n<2$ holds true for all positive integers.

Given information :

The statement if $\left\{a_n\right\}$ is the sequence $\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}},\dots$ then $a_n<2$

Formula used :

Follow the steps to prove that the statement is true for any positive integer.

Anchor step: Prove that any statement $P_n$ is true for $n=1$ .

Inductive hypothesis: Assume $P_n$ is true for $n=k$ .

Inductive Step: Prove that $P_n$ is true for $n=k+1$ .

Proof :

Conjecture:

Take $P_n$ to be statement $P_n:a_n<2$

For $n=1$ ,

  $a_1=\sqrt{2}$

For $n=2$ ,

  $a_2=\sqrt{2+\sqrt{2}}$

And,

  $a_n=\sqrt{2+\sqrt{2+\dots }}$ .

Anchor step:

Prove that $P_n$ is true for $n=1$ .

For $n=1$ ,

  $\begin{array}{l}{a}_1\stackrel{?}{<}2\\ \sqrt{2}\stackrel{?}{<}2\\ \sqrt{2}=2\end{array}$

Inductive hypothesis:

Assume $P_n$ is true for $n=k$ . Replace $n$ with $k$ in $P_n$ .

  $P_k:a_k<2$

Inductive Step:

Prove that $P_n$ is true for $n=k+1$ . This means it is needed to be proved that $P_{k+1}$ must be true.

To prove that $P_{k+1}:a_{k+1}<2$

Proceed with the inductive hypothesis.

  $P_k:a_k<2$

Now,

  $\begin{array}{c}{a}_k<2\\ a_{k+1}=\sqrt{2+a_k}\end{array}$

As $a_k<2$ , then $a_k+2<2+2$ .

  $\begin{array}{c}{a}_k+2<4\\ \sqrt{a_k+2}<\sqrt{4}\\ a_{k+1}<2\end{array}$

The statement is same as the statement $P_{k+1}$ , so $P_{k+1}$ is true.

So, if $P_n$ is true for $n=k$ , then it is also true for $n=k+1$ .

Conclusion:

As $P_n$ is true for $n=1$ and $P_k$ further implies $P_{k+1}$ , this means $P_n$ is true for $n=2$ , $n=3$ and so on.

By mathematical induction, $P_n:a_n<2$ is true for all positive integers $n$ .