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Thread: Induction

  1. #1
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    Induction

    I've got stuck on an induction problem, any help would be great
    A(1)=√5 A(n+1)=√5*a(n)

    The terms of the sequence are: √5, √(5√5), √(5(√5(√5))),....

    Use induction to show that for natural number n, 0<A(n)<5
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  2. #2
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    Quote Originally Posted by iwish123 View Post
    I've got stuck on an induction problem, any help would be great
    A(1)=√5 A(n+1)=√5*a(n)

    The terms of the sequence are: √5, √(5√5), √(5(√5(√5))),....

    Use induction to show that for natural number n, 0<A(n)<5
    Hi iwish123,

    $\displaystyle A_1=\sqrt{5}$

    $\displaystyle A_{n+1}=\sqrt{5A_n}$

    Show by induction that $\displaystyle 0<A_n<5$

    If $\displaystyle A_n\ <\ 5$

    then $\displaystyle \sqrt{A_n}\ <\ \sqrt{5}$

    hence $\displaystyle A_{n+1}=\sqrt{5A_n}=\sqrt{5}\sqrt{A_n}\ <\ \sqrt{5}\sqrt{5}\ <\ 5$

    As $\displaystyle A_1=\sqrt{5}$

    then

    $\displaystyle A_2<5,\ A_3<5,\ A_4<5.......$

    So, $\displaystyle A_1$ being < 5 causes all subsequent terms of the sequence to be < 5 also.


    A non-inductive proof is as follows...


    $\displaystyle A_1=\sqrt{5}$

    $\displaystyle A_2=\sqrt{5\sqrt{5}}=\sqrt{5}\sqrt{\sqrt{5}}=5^{\f rac{1}{2}}5^{\frac{1}{4}}$

    $\displaystyle A_3=\sqrt{5\sqrt{5\sqrt{5}}}=\sqrt{5}\sqrt{\sqrt{5 }}\sqrt{\sqrt{\sqrt{5}}}=5^{\frac{1}{2}}5^{\frac{1 }{4}}5^{\frac{1}{8}}$

    $\displaystyle A_n=5^{\frac{1}{2}}5^{\frac{1}{4}}5^{\frac{1}{8}}. .....$

    $\displaystyle =5^{\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\fra c{1}{16}+....\right)}=5^{\frac{1}{2}\left(1+\frac{ 1}{2}+\frac{1}{4}+\frac{1}{8}+....\right)}$

    The index is a geometric series, first term=1, common ratio = 0.5 (for the part in brackets)

    $\displaystyle S_{\infty}=\frac{1}{1-\frac{1}{2}}=2$

    so the index of 5 is 1.

    Hence $\displaystyle A_n$ cannot reach 5 as a finite number of terms will not reach the sum to infinity as all the terms are positive.
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  3. #3
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    Thanks, thats a great help

    How would I go about proving that the sequence is convergent, through showing its an increasing sequence and bounded? I assume that I can say it is bounded as its been shown that the sequence is less than 5.
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  4. #4
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    Quote Originally Posted by iwish123 View Post
    Thanks, thats a great help

    How would I go about proving that the sequence is convergent, through showing its an increasing sequence and bounded? I assume that I can say it is bounded as its been shown that the sequence is less than 5.
    Yes,

    either proof indicates that $\displaystyle A_n$ is positive and always < 5,
    hence the sequence converges on 5 as shown in the non-inductive proof.
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