enter image description here goes on till infinity.

I get two solutions by rewriting the term in the form of the equation $x = 3-(2/x)$, which are $1$ and $2$.

But in my opinion this term should have only one possible value. Then which one is wrong and why?

  • 1
    $\begingroup$ Maybe it doesn't converge, but instead alternates around two poles $1$ and $2$? $\endgroup$ – Colm Bhandal Aug 6 '15 at 16:19
  • $\begingroup$ Maybe, but I haven't ever seen such cases. Can you please give me another example, where the final value keeps "jumping" among certain fixed values? $\endgroup$ – Phoenix Aug 6 '15 at 16:26
  • $\begingroup$ Seems like the answers cleared that up :) $\endgroup$ – Colm Bhandal Aug 6 '15 at 16:41

If you continue adding numbers to the expression one at a time, then you have the sequence $$ 3,\; 3-2,\; 3-\frac{2}{3},\; 3-\frac{2}{3-2},\; 3-\frac{2}{3-\frac{2}{3}},\; 3-\frac{2}{3-\frac{2}{3-2}},\;3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}\ldots, $$ or $$ 3,\; 1,\; \frac{7}{3},\; 1,\; \frac{15}{7},\; 1,\;\frac{31}{15},\;\ldots, $$ which consists of two alternating subsequences: one is identically $1$, and the other converges to $2$. Depending on exactly how you define the value of the infinite fraction (i.e., what sequence you define it to be the limit of), it could be $1$, or $2$, or non-convergent.

  • $\begingroup$ You might be interested in some other aspects I mention in my answer. Regards, $\endgroup$ – Markus Scheuer Aug 7 '15 at 15:00

Look at it as two different series and you will understand why both 1 and 2 are possible solutions of this:

series 1: $\{3-2, 3-\frac{2}{3-2}, 3-\frac{2}{3-\frac{2}{3-2}}, ...\}$
series 2: $\{3-\frac{2}{3}, 3-\frac{2}{3-\frac{2}{3}}, 3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}, ...\}$.

series 1 converges to 1 whereas series 2 converges to 2.


Note: OP's expression is usually regarded as continued fraction. We will show it has the single solution $2$.

Before we analyse OPs expression, let's have a look at a continued fraction representation of $\sqrt{2}$

\begin{align*} \sqrt{2}=[1;2,2,2,\ldots]=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ddots}}} \end{align*}

The convenient notation $[1;2,2,2,\ldots]$ shows the integer part $1$ of $\sqrt{2}$ in the first position, followed by the successive values $2$ in the denominators left from the '$+$' sign.

Since the numerator is always $1$ it's called a simple continued fraction. In general a simple continued fraction can be written as

\begin{align*} x=[a_0;a_1,a_2,a_3,\ldots]=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\ddots}}} \end{align*}

The approximation by finite continued fractions of $x$ is the sequence \begin{align*} [a_0],[a_0;a_1],[a_0;a_1,a_2],[a_0;a_1,a_2,a_3],\ldots \end{align*}

In case of $\sqrt{2}$ we obtain \begin{align*} [a_0]&=[1]=1\\ [a_0;a_1]&=[1;2]=1+\frac{1}{2}=\frac{3}{2}\\ [a_0;a_1,a_2]&=[1;2,2]=1+\frac{1}{2+\frac{1}{2}}=\frac{7}{5}\\ [a_0;a_1,a_2,a_3]&=[1;2,2,2]=1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}\\ &\ldots \end{align*}

In general we consider a continued fraction in the form

\begin{align*} x=a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{a_3+\ddots}}}\tag{1} \end{align*}

and the approximation of $x$ is analogously

\begin{align*} x_0&=a_0\\ x_1&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.}=a_0+\frac{b_1}{a_1}\\ x_2&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.} +\frac{\left.b_2\right|}{\left|a_2\right.} =a_0+\frac{b_1}{a_1+\frac{b_2}{a_2}}\\ x_3&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.} +\frac{\left.b_2\right|}{\left|a_2\right.} +\frac{\left.b_3\right|}{\left|a_3\right.} =a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{a_3}}}\\ &\ldots \end{align*}

Now we are well prepared to take a look at OPs expression

\begin{align*} x=3+\frac{-2}{3+\frac{-2}{3+\frac{-2}{3+\ddots}}} \end{align*}

Here we put the '$-$' sign to the denominator in order to get the same representation as in (1). In a more compact way we can write

\begin{align*} x=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\cdots \end{align*}

The approximation of $x$ by its finite continued fractions is

\begin{align*} x_0&=3\\ x_1&=3+\frac{\left.-2\right|}{\left|3\right.}=3+\frac{-2}{3}=\frac{7}{3}\\ x_2&=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} =3+\frac{-2}{3}=\frac{15}{7}\\ x_3&=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} =3+\frac{-2}{3+\frac{-2}{3+\frac{-2}{3}}}=\frac{31}{15}\tag{2}\\ &\ldots \end{align*}

We see from (2) that $x$ converges to $\lim_{n\rightarrow \infty}\left(2+\frac{1}{n}\right)=2$ which is easily to prove.

Conclusion: The infinite continued fraction $x$ is defined as the limit of the corresponding approximating finite continued fractions in (2) and we so obtain $x=2$ as the only solution.

Epilog: Another nice gem is the continued fraction of $e$

\begin{align*} e=[2;1,2,1,1,4,1,1,6,1,1,8,...] \end{align*}

and here is a short proof of it.


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