Help in proving sequence by induction

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Homework Statement


The sequence [tex]u_1,u_2,u_3,...[/tex] is such that [tex]u_1=1[/tex] and [tex]u_{n+1}=-1+{\sqrt{u_n+7}}[/tex]

a) Prove by induction that [tex]u_n<2 for all n\geq1[/tex]
b) show that if [tex]u_n=2-\epsilon[/tex], where [tex]\epsilon[/tex] is small, then [tex]u_{n+1}\approx 2-\frac{1}{6}\epsilon[/tex]



Homework Equations



The Attempt at a Solution


[tex]u_{n+1}=-1+sqrt{u_n+7}[/tex]

[tex]\Rightarrow u_n=(u_{n+1}+1)^2-7[/tex]

Assume statement is true for all [tex]k\geq1[/tex]
then [tex]u_k<2[/tex]
[tex]\Rightarrow (u_{k+1}+1)^2-7<2[/tex]


[tex](u_{k+1}+1)^2-(3)^2<0[/tex]

[tex]((u_{k+1}+1)-3)((u_{k+1}+1)-3)<0[/tex]

[tex](u_{k+1}+1)-3>0[/tex] AND [tex](u_{k+1}+1)-3<0[/tex]
[tex] u_{k+1}+1>3<br /> <br /> u_{k+1}>2[/tex]

Thus [tex]u_{n+1}>2[/tex] is true

[tex] (u_{k+1}+1)-3<0<br /> <br /> u_{k+1}+1<-3<br /> <br /> u_{k+1}<-2[/tex]
does this affect anything in my proof?

I didn't bother to substitute the values of [tex]u_1[/tex] and [tex]u_2[/tex] and so forth as i have already done it and it is so for all [tex]n\geq1[/tex]

but I do not know how to do part b)
 
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rock.freak667 said:

Homework Statement


The sequence [tex]u_1,u_2,u_3,...[/tex] is such that [tex]u_1=1[/tex] and [tex]u_{n+1}=-1+{\sqrt{u_n+7}}[/tex]

a) Prove by induction that [tex]u_n<2 for all n\geq1[/tex]
b) show that if [tex]u_n=2-\epsilon[/tex], where [tex]\epsilon[/tex] is small, then [tex]u_{n+1}\approx 2-\frac{1}{6}\epsilon<br /> <br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <h2>The Attempt at a Solution</h2><br /> [tex]u_{n+1}=-1+sqrt{u_n+7}[/tex]<br /> <br /> [tex]\Rightarrow u_n=(u_{n+1}+1)^2-7[/tex]<br /> <br /> Assume statement is true for all [tex]k\geq1[/tex]<br /> [/tex]
[tex] <br /> Didn't you mean assume it's true for [tex]n\leq{k}[/tex] ? Actually assuming it is true for n=k is sufficient. But either way is fine.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> then [tex]u_k<2[/tex]<br /> [tex]\Rightarrow (u_{k+1}+1)^2-7<2[/tex]<br /> <br /> <br /> [tex](u_{k+1}+1)^2-(3)^2<0[/tex] </div> </div> </blockquote><br /> Good so far... but at this point I'd take the 3^2 to the other side... if x^2<a^2 where a>0, then x<a and x>-a... and then you can use the x<a part to finish...[/tex]
 
rock.freak667 said:
So then expand [tex]\sqrt{u_n+7}[/tex] using taylor series and it should work out then?

You should substitute in [tex]2-\epsilon[/tex] for [tex]u_n[/tex] in the expression for [tex]u_{n+1}[/tex]... then get the taylor series of [tex]u_{n+1}[/tex] in terms of epsilon...