MATH220 April 2011
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Question 07 (c)

Let $\{b_{n}\}$ be a sequence defined by
 $b_{1}=2\quad {\text{ and }}\quad b_{n+1}={\frac {b_{n}+{\sqrt {b_{n}}}}{2}}\quad {\text{ for each }}n\in \mathbb {N} .$
Prove that
 $b_{n+1}\leq b_{n}\quad {\text{ for each }}n\in \mathbb {N} .$

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Solution

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We will prove this statement by induction on $n$.
First, for $n=1$. By definition of the sequence we have that
 $\displaystyle {\begin{aligned}b_{2}&={\frac {b_{1}+{\sqrt {b_{1}}}}{2}}\\&={\frac {2+{\sqrt {2}}}{2}}\\&\leq {\frac {2+2}{2}}\\&=2\\&\leq b_{1}\end{aligned}}$
which proves the first step of our induction.
Let us now assume that the statement is true for all values of $n$ up to $m$ and let us show that the statement holds for $n=m+1$.
In part (b) we showed that
 $\displaystyle b_{n}\geq 1\quad {\text{for all }}n\in \mathbb {N}$
hence we can guarantee that
 $\displaystyle {\sqrt {b_{n}}}\leq b_{n}\quad {\text{for all }}n\in \mathbb {N}$
and so, we have that
 $\displaystyle {\begin{aligned}b_{m+1}&={\frac {b_{m}+{\sqrt {b_{m}}}}{2}}\\&\leq {\frac {b_{m}+b_{m}}{2}}\\&=b_{m}\end{aligned}}$
which concludes our proof.

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