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I know how to calculate the exact value for continued fractions such as


However, is it possible to find the value of continued fraction




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Certainly this continued fraction will converge very quickly to its limit; you should try writing a short program to compute the first 30 or so convergents. If you are asking whether this continued fraction has a "nice" expression, the answer is that it depends on what you mean by "nice." Continued fractions may not represent algebraic numbers like the first example; $\pi$ and $e$ have well-known and quite regular continued fraction representations, but both are transcendental numbers. –  neuguy Jul 3 '13 at 23:33
Algebraic numbers have convergents (the truncations of the continued fractions) converging 'rather slowly' to the limit. This is likely to not be an algebraic number, i.e. it will be transcendental. –  ABC Jul 4 '13 at 0:17
@proximal the continued fractions of $\pi$ and $e$ don't seem to be quite regular. You may be thinking of other expressions looking similar to continued fractions. –  ABC Jul 4 '13 at 0:21
@franklin.vp $e$ apparently has a regular pattern, in the sense that it goes $1,n+2,1$ in the 'repeated' part. So it starts $[2,1,2,1,1,4,1,1,6,1,1,8,1]$. In any case, it converges after 8! to $1.461 783 355 000 579 602 560 079 367 397$ –  wendy.krieger Jul 4 '13 at 9:03
In general, if you truncate $\beta = [a_0; a_1, a_2\ldots]$ to a finite fraction $[a_0; a_1, \ldots, a_n]$ you get a good rational approximation to $\beta$, and the quality of the approximation is better if the remaining $a_i$ are larger integers. For example, the excellence of the approximation $\pi \approx\frac{355}{133}$ is directly related to the fact that it is the truncation of $\pi$'s continued fraction $[3; 7, 15, 1, 292, 1,1,1]$ just after the suprisingly large integer 292. Your fraction has even larger partial denominators, so it is easy to find a good rational approximation to it. –  MJD Jul 6 '13 at 4:03

1 Answer 1

It converges, but not to anything particular at all.

Take a look at this

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