Science:Math Exam Resources/Courses/MATH220/April 2011/Question 07 (b)

MATH220 April 2011

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Question 07 (b)
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 $1\leq b_{n}\leq 2\quad {\text{ for all }}n\in \mathbb {N} .$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use induction!

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We will prove this by induction on $n$. First, for $n=1$, since $b_{1}=2$ we clearly have that $\displaystyle 1\leq b_{1}\leq 2$ Now, let us assume that the statement is true up to $n=m$, that is $\displaystyle 1\leq b_{n}\leq 2\quad {\text{ for }}n=1,\dots ,m$ and let us show that $\displaystyle 1\leq b_{m+1}\leq 2$ By definition of the sequence ${b_{n}}$ we have that $b_{m+1}={\frac {b_{m}+{\sqrt {b_{m}}}}{2}}$ Since $b_{m}$ ≤ 2 we have that $\displaystyle {\sqrt {b_{m}}}\leq {\sqrt {2}}\leq 2$ and so $\displaystyle b_{m+1}={\frac {b_{m}+{\sqrt {b_{m}}}}{2}}\leq {\frac {2+2}{2}}=2$ And since $b_{m}$ ≥ 1 we have that $\displaystyle {\sqrt {b_{m}}}\geq 1$ and so $\displaystyle b_{m+1}={\frac {b_{m}+{\sqrt {b_{m}}}}{2}}\geq {\frac {1+1}{2}}=1$ These two arguments show that $\displaystyle 1\leq b_{m+1}\leq 2$ and hence conclude our proof.

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