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What is the continued fraction for $\displaystyle\sum_{i=1}^n\frac{1}{2^{2^i}}$

It seems to be "almost" periodic, but I can't figure out the exact way to express it.

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    $\begingroup$ If the sum is started at $i = 0$, an explicit recurrence is given in oeis.org/classic/A007400 . The sum in this question is oeis.org/classic/A006464 , no formula is given there but perhaps a similar recurrence holds. $\endgroup$ – Rahul Oct 14 '10 at 8:09
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We can apply the following general transformation formula of a series into a continued fraction, which one can justify (see Addendum 1 and Addendum 2) by comparing the continued fraction fundamental recurrence relations with the series partial sums recurrence:

$$\sum_{n=1}^{N}\dfrac{u_{n}}{v_{n}}=\dfrac{u_{1}}{v_{1}+\underset{n=1}{ \overset{N-1}{\mathbb{K}}}\left(\left( -\dfrac{u_{n+1}}{u_{n}}v_{n}^{2}\right) /\left( v_{n+1}+\dfrac{u_{n+1}}{u_{n}}v_{n}\right)\right) }.$$

In this case, we have $u_{n}=1$, $v_{n}=2^{\left( 2^{n}\right) }$:

$$\sum_{n=1}^{N}\dfrac{1}{v_{n}}=\dfrac{1}{4+\underset{n=1}{\overset{N-1}{ \mathbb{K}}}\left( \left( -v_{n}^{2}\right) /\left( v_{n+1}+v_{n}\right)\right) }$$

$$\sum_{n=1}^{N}\dfrac{1}{2^{2^{n}}}=\dfrac{1}{4+\underset{n=1}{\overset{N-1}{\mathbb{K}}}\left(\left( -2^{2^{n+1}}\right)/\left( 2^{2^{n+1}}+2^{2^{n}}\right)\right) }$$

$$=\dfrac{1}{4+}\dfrac{-16}{20+}\cdots \dfrac{-2^{2^{n+1}}}{ 2^{2^{n+1}}+2^{2^{n}}+}{\cdots }\dfrac{-2^{2^{N}}}{ 2^{2^{N}}+2^{2^{N-1}}}.$$

The transformation of the series into a continued fraction is

$$\sum_{n=1}^{\infty}\dfrac{1}{2^{2^{n}}}=\dfrac{1}{4+\underset{n=1}{\overset{\infty}{\mathbb{K}}}\left(\left( -2^{2^{n+1}}\right)/\left( 2^{2^{n+1}}+2^{2^{n}}\right) \right) }.$$


Addendum 1: The series partial sums

$$s_{n}=\sum_{k=1}^{n}\frac{u_{k}}{v_{k}}=\frac{A_{n}}{B_{n}}$$

verify, for $n\geq 2$,

$$s_{n}=s_{n-1}+\frac{u_{n}}{v_{n}}=\frac{A_{n-1}}{B_{n-1}}+\frac{u_{n}}{v_{n} }=\frac{v_{n}A_{n-1}+u_{n}B_{n-1}}{v^{n}B_{n-1}}=\frac{A_{n}}{B_{n}}$$

which means that

$$A_{n}=v_{n}\;A_{n-1}+u_{n}\;B_{n-1}$$

$$B_{n}=v_{n}\;B_{n-1}.$$

The truncated continued fraction

$$\underset{k=1}{\overset{n}{\mathbb{K}}}\left( u_{k}/v_{k}\right) =\frac{ A_{n}}{B_{n}}$$

verifies:

$$A_{n}=b_{n}\;A_{n-1}+a_{n}\;A_{n-2}\qquad A_{0}=0$$

$$B_{n}=b_{n}\;B_{n-1}+a_{n}\;B_{n-2}\qquad B_{0}=1.$$


Addendum 2: Detailed algebraic computation. For $n=1$ we have

$$\frac{u_{1}}{v_{1}}=\frac{a_{1}}{b_{1}}=\frac{A_{1}}{B_{1}}\qquad u_{1}=a_{1}\qquad v_{1}=b_{1}.$$

Replacing $n-1$ for $n$ in the first recurrence we get for $n\geq 3$

$$A_{n-1}=v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}$$

$$B_{n-1}=v_{n-1}\;B_{n-2}$$

which in turn gives:

$$A_{n}=v_{n}\;A_{n-1}+u_{n}\;B_{n-1}$$

$$=v_{n}\;\left( v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}\right) +u_{n}\;\left( v_{n-1}\;B_{n-2}\right) $$

$$=v_{n}\;v_{n-1}\;A_{n-2}+\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;B_{n-2}$$

and

$$B_{n}=v_{n}\;B_{n-1}=v_{n}\;v_{n-1}\;B_{n-2}.$$

The same substitution in the second recurrence yields (for $n\geq 3$):

$$A_{n-1}=b_{n-1}\;A_{n-2}+a_{n-1}\;A_{n-3}$$

$$B_{n-1}=b_{n-1}\;B_{n-2}+a_{n-1}\;B_{n-3}.$$

Combining everything we obtain:

$$A_{n}=b_{n}\;A_{n-1}+a_{n}\;A_{n-2}$$

$$=b_{n}\;\left( v_{n-1}\;A_{n-2}+u_{n-1}\;B_{n-2}\right) +a_{n}\;A_{n-2}$$

$$=\left( b_{n}\;v_{n-1}+a_{n}\;\right) \;A_{n-2}+b_{n}\;u_{n-1}\;B_{n-2}$$

and

$$B_{n}=b_{n}\;B_{n-1}+a_{n}\;B_{n-2}$$

$$=b_{n}\;\left( v_{n-1}\;B_{n-2}\right) +a_{n}\;B_{n-2}$$

$$=\left( b_{n}\;v_{n-1}+a_{n}\right) \;B_{n-2}$$

Comparing both $A_{n}$ and $B_{n}$ formulae

$$A_{n}=v_{n}\;v_{n-1}\;A_{n-2}+\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;B_{n-2}$$

$$A_{n}=\left( b_{n}\;v_{n-1}+a_{n}\;\right) \;A_{n-2}+b_{n}\;u_{n-1}\;B_{n-2}$$

and

$$B_{n}=v_{n}\;v_{n-1}\;B_{n-2}$$

$$B_{n}=\left( b_{n}\;v_{n-1}+a_{n}\right) \;B_{n-2}$$

one concludes that

$$v_{n}\;u_{n-1}+u_{n}\;v_{n-1}=b_{n}\;u_{n-1}$$

$$v_{n}\;v_{n-1}=b_{n}\;v_{n-1}+a_{n}.$$

Hence

$$a_{n}=v_{n}\;v_{n-1}-b_{n}\;v_{n-1}$$

$$=v_{n}\;v_{n-1}-\left( v_{n}\;u_{n-1}+u_{n}\;v_{n-1}\right)\;v_{n-1}/u_{n-1}$$

$$=v_{n}\;v_{n-1}-v_{n}\;v_{n-1}-u_{n}\;v_{n-1}\;v_{n-1}/u_{n-1}$$

$$=-\frac{u_{n}}{u_{n-1}}v_{n-1}^{2},$$

and

$$b_{n}\;u_{n-1}=v_{n}\;u_{n-1}+u_{n}\;v_{n-1}$$

$$b_{n}=v_{n}+\frac{u_{n}}{u_{n-1}}v_{n-1}.$$

Thus for $n\geq 2$

$$a_{n}=-\frac{u_{n}}{u_{n-1}}v_{n-1}^{2}$$

$$b_{n}=v_{n}+\frac{u_{n}}{u_{n-1}}v_{n-1}.$$

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  • $\begingroup$ Thank you so much. I hadn't seen the relation for continued fractions and series partial sums yet. I just started working with continued fractions recently. $\endgroup$ – James Oct 21 '10 at 5:17

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