sequence problem: Suppose c > 0, x_1 = sqrt(c), and x_n+1
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First part:
\(\displaystyle \L x_1 = \sqrt{c} \\ x_{n+1} = \sqrt{c+x_{n}}\)
(1) Prove that \(\displaystyle x_n < x_{n+1}\):
Pf: Proof by indction on n.
Initial case:n=1. This is obvious.
Inductive step: Let P(n) be the assertion that x<sub>n</sub> < x<sub>n+1</sub>. Assume P(n) and show P(n)=>P(n+1).
\(\displaystyle x_{n+2} \,\, = \,\, \sqrt{c+x_{n+1}}\) by definition of x<sub>n</sub>.
Also, \(\displaystyle \sqrt{c+x_{n+1}}\,\, >\,\, \sqrt{c+x_n}\) Since x<sub>n+1</sub> > x<sub>n</sub>.
But that is exactly the definition of x<sub>n+1</sub>. So, \(\displaystyle x_{n+2}\,\, >\,\, \sqrt{c+x_{n}}\,\, = \,\,x_{n+1}\). So we have finished the proof.
\(\displaystyle \L x_1 = \sqrt{c} \\ x_{n+1} = \sqrt{c+x_{n}}\)
(1) Prove that \(\displaystyle x_n < x_{n+1}\):
Pf: Proof by indction on n.
Initial case:n=1. This is obvious.
Inductive step: Let P(n) be the assertion that x<sub>n</sub> < x<sub>n+1</sub>. Assume P(n) and show P(n)=>P(n+1).
\(\displaystyle x_{n+2} \,\, = \,\, \sqrt{c+x_{n+1}}\) by definition of x<sub>n</sub>.
Also, \(\displaystyle \sqrt{c+x_{n+1}}\,\, >\,\, \sqrt{c+x_n}\) Since x<sub>n+1</sub> > x<sub>n</sub>.
But that is exactly the definition of x<sub>n+1</sub>. So, \(\displaystyle x_{n+2}\,\, >\,\, \sqrt{c+x_{n}}\,\, = \,\,x_{n+1}\). So we have finished the proof.
pka
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The point of the first part is to show that the sequence \(\displaystyle \left( {x_n } \right)\) is both increasing and bounded. Therefore, \(\displaystyle \left( {x_n } \right)\) has a limit call it L.
Then \(\displaystyle \left( {x_n } \right) \to L\;\& \;\left( {x_{n + 1} } \right) \to L\) which gives us \(\displaystyle L = \sqrt {c + L}\)
Then \(\displaystyle \left( {x_n } \right) \to L\;\& \;\left( {x_{n + 1} } \right) \to L\) which gives us \(\displaystyle L = \sqrt {c + L}\)