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Math Help - sum to infinity

  1. #1
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    sum to infinity

    Determine the value of

     <br />
\frac{\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...\text{infinity}}}}}{\sqrt{2\sqrt{2\sqrt{2...\  text{infinity}}}}}

    My thoughts on this :

    For the numerator :

     <br />
4^{\frac{1}{3}} , 4^{\frac{1}{9}} , 4^{\frac{1}{27}} , ... <br />

    a=4^{\frac{1}{3}} , r=??

    Same goes for the deniminator :

    2^{\frac{1}{2}} , 2^{\frac{1}{4}} , 2^{\frac{1}{8}} , ...

     <br />
a=2^{\frac{1}{2}} , r=?? <br />
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Determine the value of

     <br />
\frac{\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...\text{infinity}}}}}{\sqrt{2\sqrt{2\sqrt{2...\  text{infinity}}}}}

    My thoughts on this :

    For the numerator :

     <br />
4^{\frac{1}{3}} , 4^{\frac{1}{9}} , 4^{\frac{1}{27}} , ... <br />

    a=4^{\frac{1}{3}} , r=??

    Same goes for the deniminator :

    2^{\frac{1}{2}} , 2^{\frac{1}{4}} , 2^{\frac{1}{8}} , ...

     <br />
a=2^{\frac{1}{2}} , r=?? <br />
    It should be

     4^{\frac{1}{3}} , 4^{\frac{\frac{1}{3}+1}{3}} , 4^{\frac{\frac{\frac{1}{3}+1}{3}+1}{3}}

    we can find the relation :  a_{n+1} = (4(a_n))^{\frac{1}{3}} and the second  b_{n+1} = (2(b_n))^{\frac{1}{2}}

    If the sequences converge ,  a_{n} = (4(a_n))^{\frac{1}{3}} and  b_{n+1} = (2(b_n))^{\frac{1}{2}}

    finally we can obtain  \lim_{n\to\infty} a_n   = 2
     \lim_{n\to\infty} b_n = 2 so their ratio is one
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  3. #3
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Determine the value of

     <br />
\frac{\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...\text{infinity}}}}}{\sqrt{2\sqrt{2\sqrt{2...\  text{infinity}}}}}

    My thoughts on this :

    For the numerator :

     <br />
4^{\frac{1}{3}} , 4^{\frac{1}{9}} , 4^{\frac{1}{27}} , ... <br />

    a=4^{\frac{1}{3}} , r=??

    Same goes for the deniminator :

    2^{\frac{1}{2}} , 2^{\frac{1}{4}} , 2^{\frac{1}{8}} , ...

     <br />
a=2^{\frac{1}{2}} , r=?? <br />
    Your thoughts are correct, but you're not sure how to fit them together, are you? Look at the denominator like this. First, note that:

    \sqrt{2 \times x}=\sqrt{2} \times \sqrt{x}, or (2\times x)^{\frac12}=2^{\frac12}\times x^{\frac12}

    whatever x may stand for.

    So if
    x=\sqrt{2\times y},\,\sqrt{2\times x}=2^{\frac12}\times (\sqrt{2\times y})^{\frac12} =2^{\frac12}\times(2^{\frac12}\times y^{\frac12})^{\frac12}=2^{\frac12}\times 2^{\frac14}\times y^{\frac14}

    In the same way, if y = \sqrt{2\times z}, y^{\frac14}=2^{\frac18}\times z^{\frac18}, so:

    \sqrt{2\times x} =2^{\frac12}\times 2^{\frac14}\times 2^{\frac18}\times z^{\frac18}

    ... and so on. So the denominator becomes:

     \sqrt{2\sqrt{2\sqrt{2\sqrt{...}}}}=2^{\frac12}\tim  es 2^{\frac14}\times 2^{\frac18}\times ... = 2^{\frac12+\frac14+\frac18+...}

    And we now note that \tfrac12+\tfrac14+\tfrac18+... is an infinite GP, with first term a = \tfrac12, common ratio r = \tfrac12, and with a sum to infinity of \frac{a}{1-r}=1.

    In the same way, the numerator is 4^{\frac13}\times 4^{\frac19}\times4^{\frac{1}{27}}\times ...= 4^{\frac13+\frac19+\frac{1}{27}+...}

    And \tfrac13+\tfrac19+\tfrac{1}{27}+... = \frac{\frac13}{1-\frac13} = \tfrac12

    So the original expression simply becomes \frac{4^{\frac12}}{2^1}=1

    Grandad
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  4. #4
    Moo
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    Hello,

    Here is another way...

    Let a=\frac{\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}}{\sqrt{2\sqrt{2\sqrt{2...}}}}

    we have a cube root and a square root. so put everything to the power 6 :

    \begin{aligned}<br />
a^6 &=\tfrac{4^2 \left(\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}\right)^2}{2^3 \left(\sqrt{2\sqrt{2\sqrt{2...}}}\right)^3} \\<br />
&=2\cdot \tfrac{\left(\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}\right)^2}{\left(\sqrt{2\sqrt{2\sqrt{2...}  }}\right)^2 \left(\sqrt{2\sqrt{2\sqrt{2...}}}\right)} \\<br />
&=2 \cdot\tfrac{\left(\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}\right)^2}{(2 \sqrt{2\sqrt{2\sqrt{2...}}}) (\sqrt{2\sqrt{2\sqrt{2...}}})} \\<br />
\end{aligned}

    \begin{aligned}<br />
a^6 &=\tfrac{\left(\sqrt[3]{4\sqrt[3]{4\sqrt[3]{4...}}}\right)^2}{\left(\sqrt{2\sqrt{2\sqrt{2...}  }}\right)^2} \\<br />
&=a^2 \end{aligned}

    Hence a^6-a^2=0 \Rightarrow a^2(a-1)(a+1)(a^2+1)=0

    a is not 0 (hmmm we'll have to justify it), a is positive, and real.
    so a=1...
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  5. #5
    Senior Member
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    Quote Originally Posted by Grandad View Post
    Hello thereddevilsYour thoughts are correct, but you're not sure how to fit them together, are you? Look at the denominator like this. First, note that:

    \sqrt{2 \times x}=\sqrt{2} \times \sqrt{x}, or (2\times x)^{\frac12}=2^{\frac12}\times x^{\frac12}

    whatever x may stand for.

    So if x=\sqrt{2\times y},\,\sqrt{2\times x}=2^{\frac12}\times (\sqrt{2\times y})^{\frac12} =2^{\frac12}\times(2^{\frac12}\times y^{\frac12})^{\frac12}=2^{\frac12}\times 2^{\frac14}\times y^{\frac14}

    In the same way, if y = \sqrt{2\times z}, y^{\frac14}=2^{\frac18}\times z^{\frac18}, so:

    \sqrt{2\times x} =2^{\frac12}\times 2^{\frac14}\times 2^{\frac18}\times z^{\frac18}

    ... and so on. So the denominator becomes:

     \sqrt{2\sqrt{2\sqrt{2\sqrt{...}}}}=2^{\frac12}\tim  es 2^{\frac14}\times 2^{\frac18}\times ... = 2^{\frac12+\frac14+\frac18+...}

    And we now note that \tfrac12+\tfrac14+\tfrac18+... is an infinite GP, with first term a = \tfrac12, common ratio r = \tfrac12, and with a sum to infinity of \frac{a}{1-r}=1.

    In the same way, the numerator is 4^{\frac13}\times 4^{\frac19}\times4^{\frac{1}{27}}\times ...= 4^{\frac13+\frac19+\frac{1}{27}+...}

    And \tfrac13+\tfrac19+\tfrac{1}{27}+... = \frac{\frac13}{1-\frac13} = \tfrac12

    So the original expression simply becomes \frac{4^{\frac12}}{2^1}=1

    Grandad
    Thanks Grandad , you really understand the problems a student would possibly face ..


    Thanks Moo too , for another interesting solution ..
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