# Math Help - sum to infinity

1. ## sum to infinity

Determine the value of

$
\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 :

$
4^{\frac{1}{3}} , 4^{\frac{1}{9}} , 4^{\frac{1}{27}} , ...
$

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

Same goes for the deniminator :

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

$
a=2^{\frac{1}{2}} , r=??
$

2. Originally Posted by thereddevils
Determine the value of

$
\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 :

$
4^{\frac{1}{3}} , 4^{\frac{1}{9}} , 4^{\frac{1}{27}} , ...
$

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

Same goes for the deniminator :

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

$
a=2^{\frac{1}{2}} , r=??
$
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

3. Hello thereddevils
Originally Posted by thereddevils
Determine the value of

$
\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 :

$
4^{\frac{1}{3}} , 4^{\frac{1}{9}} , 4^{\frac{1}{27}} , ...
$

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

Same goes for the deniminator :

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

$
a=2^{\frac{1}{2}} , r=??
$
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 $\frac{4^{\frac12}}{2^1}=1$

4. 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}
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} \\
&=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)} \\
&=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...}}})} \\
\end{aligned}

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

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 $\frac{4^{\frac12}}{2^1}=1$