[Math] Writing continued fractions of irrational numbers as infinite series

continued-fractionsirrational-numberssequences-and-series

Infinite sums have been formulated for famous irrational numbers, such as $\pi, \phi,e,\sqrt2$ and a few others that can be listed here and here:

Here are some examples: (There are more examples if you follow the links above)

$$\sqrt 2 = \sum_{k=0}^\infty{\frac{(2k+1)!}{2^{3k+1} (k!)^2}}$$
$$\pi = \sum_{k=0}^\infty{\frac{(2^{k+1})(k!)^2}{2k+1}}$$
$$\phi = \frac{13}{8} + \sum_{k=0}^\infty{\frac{((-1)^{k+1})(2k+1)!}{((k+2)!)(k!)(4^{2k+3})}}$$
$$e = \sum_{k=0}^\infty{\frac{1}{k!}}$$

In the section that follows I will just be discussing non-transcendental numbers (such as $\sqrt2,\sqrt3,\sqrt5$, etc (not $\pi,e$, etc)

As far as I know there are no known infinite sums for $\sqrt3,\sqrt5$ and irrational numbers such which can be written as $\sqrt[a]b$ Where $a$ is an real number > 0, and b is a prime number or the product of prime numbers. And such that $\sqrt[a]b$ is not an integer (perfect squares or perfect cubes).

However, there are continued fractions which can represent some of these numbers, such as $\sqrt5$:

$$\sqrt5 = 2+ \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + …}}}}}}$$

This wikipedia section discusses this concept nicely: Click here

SO MY QUESTION IS:

If there is a continued fraction for $\sqrt x$, then how do we go from the continued fraction to an infinite sum? (Conversion maybe?) Let's take $\sqrt5$ as an example:

How I thought of doing this, would to work out rational approximations by taking the continued fraction to certain terms (T), such as:

  • T1 = $2 = \frac{2}{1}$
  • T2 = $2+ \frac{1}{4} = \frac{9}{4} = 2,25$
  • T3 = $2+ \frac{1}{4 + \frac{1}{4}} = \frac{38}{17} = 2,234294118…$
  • T4 = $2+ \frac{1}{4 + \frac{1}{4+\frac{1}{4}}} = \frac{161}{72} = 2,236111111…$

  • T5 = $2+ \frac{1}{4 + \frac{1}{4+\frac{1}{4+\frac{1}{4}}}} = \frac{682}{305} = 2,236065574…$

  • T(n) = $\frac{T_{n+1}(OEIS: A001077)}{T_{n+1}(OEIS: A001076)}$

and the list goes on (if we did this "infinitely" many times we should theoretically reach $\sqrt5$ and that is what we are looking for 🙂

Then what I tried to do is take these rational approximations and finding the difference between them, so we could then write it as an infinite sum, comprising of adding the differences (D) of these rational approximations.

  • $D1 = T2-T1 = \frac{9}{4} – 2 = \frac{1}{4}$
  • $D2 = T3-T2 = \frac{38}{17} – \frac{9}{4} = -\frac{1}{68}$
  • $D3 = T4-T3 = \frac{161}{72} – \frac{38}{17} = \frac{1}{1224}$
  • $D4 = T5-T4 = \frac{682}{305} – \frac{161}{72} = -\frac{1}{21960}$

The numerator stays 1 and the denominators continue in the following sequence: https://oeis.org/A156084 (Each time the term alternates between being positive and negative). Let's call the absolute value of each term in the sequence $a_k$, $(a_1 = \frac{1}{4},a_2 = \frac{1}{68})$; then$\sqrt5$ could be written as:

$$\sqrt5 = 2 + \sum_{k=0}^\infty{(-1)^k} . {a_{k+1}} = 2 + \frac{1}{4} – \frac{1}{68} + \frac{1}{1224} – \frac{1}{21960} …$$

Kind Regards

Joshua

Best Answer

You can use the convergents to the continued fraction to obtain series, since the convergents give close rational approximations to $5$. $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$

Using $\sqrt{5} \approx \lfrac{9}{4}$ we get:

$\sqrt{5} = \lfrac94 ( 1 - \lfrac1{9^2} )^\lfrac12 = \lfrac{9}{4} \sum_{k=0}^\infty \binom{1/2}{k} \lfrac{(-1)^k}{9^{2k}} = \lfrac{9}{4} - \lfrac{9}{4} \sum_{k=1}^\infty \lfrac{2 \cdot (2k-2)!}{k! \cdot (k-1)! \cdot 18^{2k}}$.

Using $\sqrt{5} \approx \frac{38}{17}$ we get:

$\sqrt{5} = \lfrac{38}{17} ( 1 + \lfrac1{38^2} )^\lfrac12 = \lfrac{38}{17} \sum_{k=0}^\infty \binom{1/2}{k} \lfrac{1}{38^{2k}} = \lfrac{38}{17} - \lfrac{38}{17} \sum_{k=1}^\infty \lfrac{(-1)^k \cdot 2 \cdot (2k-2)!}{k! \cdot (k-1)! \cdot 76^{2k}}$.

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