Find the value of ​

Hmm.

.

This is simply "continued fraction" of: 8/11, which is your answer.

Logic Melody? What logic. This has been around since L. Euler!!.

Start at the bottom with 2. Take its reciprocal =1/2 + 1 to the left=1 1/2. Multiply by the same 2 at the bottom =3. Take the last mixed fraction we just got, i.e., 1 1/2. Take the reciprocal of this =2/3 + 2 to its left=2 2/3. Multiply this by the last whole integer we got, which was a 3, and you get=8. Since this is 2nd to the last fraction, then 8 is the numerator. Finally, take the last mixed fraction we got, or 2 2/3 and take its reciprocal =.375 + 1 to its left=1 3/8 x last whole integer we got,8, which was the numerator and you get: 1/ 3/8 x 8 =11 which is your denominator. Therefore this "continued fraction" =8/11. Fair?.

Sorry, I forgot to mention that you can always enter the the left  numbers [0; 1,2,1,2] as "continued fraction" in W/A engine like this:                                                                                                   http://www.wolframalpha.com/input/?i=continued+fraction%5B0;1,2,1,2%5D.

And will always give the fraction.

Hi Alan

I have been toying this question,  I have only just worked out the relevance of the 2 answers :D

Silly me :)

Here is Alan's earlier answer:

http://web2.0calc.com/questions/find-the-value-of_3

INFINITE CONTINUED FRACTION 

\( \begin{align} Let\\ x&=\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\\ so\\ x&=\cfrac{1}{1 + \cfrac{1}{2 +x}}\\~\\ x&=\cfrac{1}{\cfrac{2+x}{2+x} + \cfrac{1}{2 +x}}\\~\\ x&=\cfrac{1}{\frac{3+x}{2+x} }\\~\\ x&=\cfrac{2+x}{3+x}\\~\\ 3x+x^2&=2+x\\ x^2+2x-2&=0\\ x&=\frac{-2\pm\sqrt{4+8}}{2}\\ x&=\frac{-2\pm2\sqrt{3}}{2}\\ x&=-1\pm\sqrt{3}\\ \end{align}\)

BUT it can be seen that x is positive so the only valid solutions is \(x=\sqrt3-1\)

SO

\(\boxed{\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\\~\\ =\sqrt3-1}\)

WHEREAS IF IT IS NOT AN INFINITE CONTINUED FRACTION THEN

\(\boxed{\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 }}}}\\~\\ =\frac{8}{11}}\)

LaTex discussion on displaying fractions :)

Heureka, I thought maybe you would like to add to this post :/     

I have also been looking at the Latex that is used in this question.

I found this reference to using Latex for fractions but i will admit i have not properly digested it yet.

http://tex.stackexchange.com/questions/59747/proper-display-of-fractions

This is the question:

\(\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\)

And this is the command fot it.

\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}

I have not used   \cfrac{}{}   before.

On that reference site there is also a \dfrac{}{} used.

If anyone would like to discuss how these are used that would because I haven't fully comprehended what they are about.