Science:Math Exam Resources/Courses/MATH220/April 2011/Question 07 (c)/Solution 1

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.