# Is $\{1,1,2,3,4,5,\cdots,i,\cdots \}$ the simple continued fraction algebraic or transcendental?

Is $$1+\cfrac{1}{1+\cfrac{1}{2+\cdots}}$$ or$\{1,1,2,3,4,5,\cdots,i,\cdots \} , i\in \mathbb{N}$ the simple continued fraction algebraic or transcendental?

Any reference is appreciated

EDIT and replacing $\{1,1,2,3,4,5,\cdots,i,\cdots \}$ with $\{1,2,2^2,2^3,2^4,2^5,\cdots,2^i,\cdots \}$ $i\in \mathbb{N}$ or any other patternful sequences that are upper unbounded ,can we get an algebraic number?

## 1 Answer

I believe this value is known to be transcendental; its value is known explicitly to be $1+\frac{I_1(2)}{I_0(2)}$, where $I_0()$ and $I_1()$ are the modified Bessel functions. For more details, see http://mathworld.wolfram.com/ContinuedFractionConstant.html . This continued fraction (and continued fractions with arithmetic progressions of coefficients in general) are related to certain Riccati equations, but unfortunately a bit of search on those terms doesn't really turn up any good lightweight accounts of the link between the two; you'll have to hunt about a bit more on your own for that.

• Steven, as to the OP's earlier question today, are there any known real algebraic numbers with unbounded "elements," as Khinchin calls them? These would be the {1,1,2,3,4,5...} above – Will Jagy Jul 30 '14 at 23:58
• yes,it is relevant to my other question about upper bound of continued fraction of algebraic number – XL _at_China Jul 31 '14 at 0:00
• @WillJagy As noted there, the expectation would be that almost all algebraic numbers should have unbounded CF coefficients, but just like with normal numbers I unfortunately don't know of any specific examples. I wonder if something can be said about the continued fraction expansion of Pisot numbers; hrm... – Steven Stadnicki Jul 31 '14 at 0:04
• @XL_at_China, appears we do not know, but there are many books on continued fractions en.wikipedia.org/wiki/Oskar_Perron – Will Jagy Jul 31 '14 at 0:07
• @XL_at_China, after searching online, seems an open problem. Repeat mentions for cube root of 2, people think unbounded but no proof. Also, bounded but not periodic gives something completely different; see The Markoff and Lagrange Spectra by Cusick and Flahive – Will Jagy Jul 31 '14 at 0:22