July 19

Let ${\displaystyle a_{n}={\frac {n^{2}}{n^{2}-1}}~.}$ Then ${\displaystyle {\frac {14+10{\sqrt {5}}}{19}}=[a_{2};a_{3},a_{4},a_{5},\ldots ]=a_{2}+{\cfrac {1}{a_{3}+{\cfrac {1}{a_{4}+{\cfrac {1}{a_{5}+\cdots }}}}}}}$

Why is this so, and are there any more examples of this kind ? — 79.113.219.182 (talk) 20:26, 19 July 2015 (UTC)[reply]

Not the complete answer but hopefully this will help. Let R be the value of the continued fraction. Then after some manipulation and cancellation you get

${\displaystyle {\frac {4}{3R}}={\cfrac {1}{1+{\cfrac {\frac {2}{3}}{1+{\cfrac {\frac {5}{6}}{1+{\cfrac {\frac {9}{10}}{1+{\cfrac {\frac {14}{15}}{1+{}\ddots }}}}}}}}}}={\cfrac {1}{1+{\cfrac {\frac {1\cdot 4}{2\cdot 3}}{1+{\cfrac {\frac {2\cdot 5}{3\cdot 4}}{1+{\cfrac {\frac {3\cdot 6}{4\cdot 5}}{1+{\cfrac {\frac {4\cdot 7}{5\cdot 6}}{1+{}\ddots }}}}}}}}}}}$

which turns out to be Gauss's continued fraction

${\displaystyle {\frac {{}_{2}F_{1}(a+1,b;c+1;z)}{{}_{2}F_{1}(a,b;c;z)}}={\cfrac {1}{1+{\cfrac {{\frac {(a-c)b}{c(c+1)}}z}{1+{\cfrac {{\frac {(b-c-1)(a+1)}{(c+1)(c+2)}}z}{1+{\cfrac {{\frac {(a-c-1)(b+1)}{(c+2)(c+3)}}z}{1+{\cfrac {{\frac {(b-c-2)(a+2)}{(c+3)(c+4)}}z}{1+{}\ddots }}}}}}}}}}}$

with a=3/2, b = 2, c = 2, z = -4. So if you can evaluate

${\displaystyle {\frac {{}_{2}F_{1}(5/2,2;3;-4)}{{}_{2}F_{1}(3/2,2;2;-4)}}}$

then you're done. There are dozens of identities and special values known for hypergeometric functions, but I'm not all that familiar with them so I'll leave it there. As for whether there are similar examples, Gauss's continued fraction gives a whole family of them, though the values you get don't usually come out to be quadratic surds and it's rare that you can manipulate the c.f. into such a simple form. --RDBury (talk) 11:00, 20 July 2015 (UTC)[reply]

Not exactly an answer for your question, but for reference, this can be evaluated with the following

Mathematica

command:

4/3 + ContinuedFractionK[n^2/(n^2 - 1), {n, 3, \[Infinity]}]

Trying some similar sequences such as ${\displaystyle {\frac {n^{2}}{n^{2}-2}}}$, ${\displaystyle {\frac {n^{2}+1}{n^{2}-1}}}$ or ${\displaystyle {\frac {n^{3}}{n^{3}-1}}}$ results in something Mathematica cannot evaluate, and likely, that does not have a simple closed form. -- Meni Rosenfeld (talk) 17:36, 20 July 2015 (UTC)[reply]

I worked out the values of the hypergeometric series and got

${\displaystyle {}_{2}F_{1}(3/2,2;2;z)={}_{1}F_{0}(3/2;;z)=(1-z)^{-{3 \over 2}}}$

which is really just the binomial theorem, and

${\displaystyle {}_{2}F_{1}(5/2,2;3;z)={\frac {{4 \over 3}(1-z)^{-{3 \over 2}}-4(1-z)^{-{1 \over 2}}+{8 \over 3}}{z^{2}}}}$

which is slightly trickier. Basically, multiply by z², then differentiate so get something similar to a binomial series for which you can get a closed expression. Use freshman calculus to integrate, put z=0 to get the constant of integration, and divide by z² to get (after two and a half sheets of scrap paper) the above expression. The quotient is then

${\displaystyle Q(z)={\frac {{}_{2}F_{1}(5/2,2;3;z)}{{}_{2}F_{1}(3/2,2;2;z)}}={\frac {{8 \over 3}((1-z)^{3 \over 2}-1)+4z}{z^{2}}}}$

and by Gauss's c.f. this is

${\displaystyle Q(z)={\cfrac {1}{1+{\cfrac {{\frac {-1\cdot 4}{4\cdot 2\cdot 3}}z}{1+{\cfrac {{\frac {-2\cdot 5}{4\cdot 3\cdot 4}}z}{1+{\cfrac {{\frac {-3\cdot 6}{4\cdot 4\cdot 5}}z}{1+{\cfrac {{\frac {-4\cdot 7}{4\cdot 5\cdot 6}}z}{1+{}\ddots }}}}}}}}}}.}$

Replace z with −4z to get

${\displaystyle Q(-4z)={\frac {((1+4z)^{3 \over 2}-1)+6z}{6z^{2}}}={\cfrac {1}{1+{\cfrac {{\frac {1\cdot 4}{2\cdot 3}}z}{1+{\cfrac {{\frac {2\cdot 5}{3\cdot 4}}z}{1+{\cfrac {{\frac {3\cdot 6}{4\cdot 5}}z}{1+{\cfrac {{\frac {4\cdot 7}{5\cdot 6}}z}{1+{}\ddots }}}}}}}}}}.}$

You can manipulate the c.f. to get something closer to the original statement

${\displaystyle {\frac {4}{3Q(-4z)}}={\frac {8z^{2}}{(1+4z)^{3 \over 2}-1-6z}}={4 \over 3}+{\cfrac {z}{{9 \over 8}+{\cfrac {z}{{16 \over 15}+{\cfrac {z}{{25 \over 24}+{\cfrac {z}{{36 \over 35}+{}\ddots }}}}}}}}.}$

Something I didn't think about earlier was that the series only converge in a neighborhood of 0, but once you have a closed form for the c.f. you can apply analytic continuation to get the two sides are equal as long as the c.f. converges. So put z = 1 in the above equation and rationalize the denominator to the original statement. --RDBury (talk) 23:28, 20 July 2015 (UTC)[reply]