Q. 86 - Calculus

Question

Consider the sequence $\left\{a_k\right\}$ defined recursively by $a1=1$ and for $k>1,a_k=\sqrt{2a_{k-1}}$.Prove that $a_k\to 2$by first proving that the limit must exist. (Hint: Use induction to show that the terms of the sequence may be expressed with the closed formula for $a_k=2^{1-{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^{k-1}})$

Step-by-Step Solution

Answer

The value of $\underset{k\to \infty }{\lim }\left(a_k\right)=2$

Step 1. Given information

Consider the given sequence $\left\{a_k\right\}$ defined recursively by $a_1=1$a nd  for $k>1,a_k=\sqrt{2a_{k-1}}$

Step 2. Finding the values of terms a 1 , a 2 , a 3

The terms of the sequence is defined as

$$\begin{gathered} a_1=1.............\left(1\right) \\ {a}_2=\sqrt{2a_1} (put k=2 in a_k=\sqrt{2a_{k-1}}) \\ {a}_2=\sqrt{2} \left(simplify\right).................\left(2\right) \end{gathered}$$

From equations,(1) and (2) ,it is observed that

$0<a_1<a_2<2 (because \sqrt{2}<2)$

Put $k=3$ in $a_k=\sqrt{2a_{k-1}}$to get

$=\sqrt{2\sqrt{2}} (substitute a_2=\sqrt{2})$

Thus $a_2<a_3<2 (because \sqrt{2\sqrt{2}}<2)$

Step 3. Finding the given sequence is bounded or not.

The general term of the sequence is defined as

$a_k=\sqrt{2\sqrt{2\sqrt{2.......(k-1) times}}}$

countinuing likewisw,the following inequality is obtained

$0<a_1<a_2<a_3<.......<a_k<2$

The sequence$\left\{a_k\right\}$ is an increasing sequence because

$0<a_1<a_2<a_3<.......<a_k$

the sequence $\left\{a_k\right\}$ has a lower bound and upper bound.Thus,the given sequence is bounded

Step 4. Find the value of lim k &#8594; &#8734; a k

The monotonic increasing sequence which is bounded above is convergent.The given sequence $\left\{a_k\right\}$ defined recursively by $a_1=1$ and for $k>1,a_k=\sqrt{2a_{k-1}}$ is increasing is bounded above by 2. thus,the sequence is convergent.

Assume the limit of the sequence $\left\{a_k\right\}$ is l.

Therefore,

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\left(\sqrt{2a_{k-1}}\right) \\ l=\sqrt{2\underset{k\to \infty }{\lim }{a}_{k-2}} (because a_k\to 1) \\ l=\sqrt{2l} \\ {l}^2-2l=0 \left(squaring\right) \\ l(l-2)=0 \left(factorize\right) \\ l=0,2 (solve for l) \end{gathered}$$

The sequence cannot converge to 0 because $a_1=1$ and the sequence is  increasing.Therefore,the value of l is 2

Thus, the value of $\underset{k\to \infty }{\lim }\left(a_k\right)=2$