Help in proving sequence by induction

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The discussion centers on proving properties of the sequence defined by u_1=1 and u_{n+1}=-1+sqrt{u_n+7}. For part a, participants discuss using induction to show that u_n<2 for all n≥1, with suggestions to manipulate the inequality and clarify assumptions. In part b, the focus shifts to approximating u_{n+1} when u_n=2-ε, with recommendations to apply Taylor series expansion for simplification. The conversation emphasizes the importance of correctly substituting values and leveraging mathematical properties to complete the proofs. Overall, the thread provides insights into tackling sequence induction and approximation techniques.
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Homework Statement


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

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



Homework Equations



The Attempt at a Solution


u_{n+1}=-1+sqrt{u_n+7}

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

Assume statement is true for all k\geq1
then u_k&lt;2
\Rightarrow (u_{k+1}+1)^2-7&lt;2


(u_{k+1}+1)^2-(3)^2&lt;0

((u_{k+1}+1)-3)((u_{k+1}+1)-3)&lt;0

(u_{k+1}+1)-3&gt;0 AND (u_{k+1}+1)-3&lt;0
<br /> u_{k+1}+1&gt;3<br /> <br /> u_{k+1}&gt;2<br />

Thus u_{n+1}&gt;2 is true

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

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

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

Homework Statement


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

a) Prove by induction that u_n&lt;2 for all n\geq1
b) show that if u_n=2-\epsilon, where \epsilon is small, then 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 /> u_{n+1}=-1+sqrt{u_n+7}<br /> <br /> \Rightarrow u_n=(u_{n+1}+1)^2-7<br /> <br /> Assume statement is true for all k\geq1<br /> <br />
<br /> <br /> Didn&#039;t you mean assume it&#039;s true for n\leq{k} ? 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 u_k&amp;lt;2<br /> \Rightarrow (u_{k+1}+1)^2-7&amp;lt;2<br /> <br /> <br /> (u_{k+1}+1)^2-(3)^2&amp;lt;0 </div> </div> </blockquote><br /> Good so far... but at this point I&#039;d take the 3^2 to the other side... if x^2&lt;a^2 where a&gt;0, then x&lt;a and x&gt;-a... and then you can use the x&lt;a part to finish...
 
For part b, use the taylor series expansion in terms of \epsilon... you only need the first two terms.
 
So then expand \sqrt{u_n+7} using taylor series and it should work out then?
 
rock.freak667 said:
So then expand \sqrt{u_n+7} using taylor series and it should work out then?

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