The series ₀F₁

The simplest case involves

${\displaystyle \,_{0}F_{1}(a;z)=1+{\frac {1}{a\,1!}}z+{\frac {1}{a(a+1)\,2!}}z^{2}+{\frac {1}{a(a+1)(a+2)\,3!}}z^{3}+\cdots .}$

Starting with the identity

${\displaystyle \,_{0}F_{1}(a-1;z)-\,_{0}F_{1}(a;z)={\frac {z}{a(a-1)}}\,_{0}F_{1}(a+1;z),}$

we may take

${\displaystyle f_{i}={}_{0}F_{1}(a+i;z),\,k_{i}={\tfrac {1}{(a+i)(a+i-1)}},}$

giving

${\displaystyle {\frac {\,_{0}F_{1}(a+1;z)}{\,_{0}F_{1}(a;z)}}={\cfrac {1}{1+{\cfrac {{\frac {1}{a(a+1)}}z}{1+{\cfrac {{\frac {1}{(a+1)(a+2)}}z}{1+{\cfrac {{\frac {1}{(a+2)(a+3)}}z}{1+{}\ddots }}}}}}}}}$

or

${\displaystyle {\frac {\,_{0}F_{1}(a+1;z)}{a\,_{0}F_{1}(a;z)}}={\cfrac {1}{a+{\cfrac {z}{(a+1)+{\cfrac {z}{(a+2)+{\cfrac {z}{(a+3)+{}\ddots }}}}}}}}.}$

This expansion converges to the meromorphic function defined by the ratio of the two convergent series (provided, of course, that a is neither zero nor a negative integer).

The series ₁F₁

The next case involves

${\displaystyle {}_{1}F_{1}(a;b;z)=1+{\frac {a}{b\,1!}}z+{\frac {a(a+1)}{b(b+1)\,2!}}z^{2}+{\frac {a(a+1)(a+2)}{b(b+1)(b+2)\,3!}}z^{3}+\cdots }$

for which the two identities

${\displaystyle \,_{1}F_{1}(a;b-1;z)-\,_{1}F_{1}(a+1;b;z)={\frac {(a-b+1)z}{b(b-1)}}\,_{1}F_{1}(a+1;b+1;z)}$

${\displaystyle \,_{1}F_{1}(a;b-1;z)-\,_{1}F_{1}(a;b;z)={\frac {az}{b(b-1)}}\,_{1}F_{1}(a+1;b+1;z)}$

are used alternately.

Let

${\displaystyle f_{0}(z)=\,_{1}F_{1}(a;b;z),}$

${\displaystyle f_{1}(z)=\,_{1}F_{1}(a+1;b+1;z),}$

${\displaystyle f_{2}(z)=\,_{1}F_{1}(a+1;b+2;z),}$

${\displaystyle f_{3}(z)=\,_{1}F_{1}(a+2;b+3;z),}$

${\displaystyle f_{4}(z)=\,_{1}F_{1}(a+2;b+4;z),}$

etc.

This gives ${\displaystyle f_{i-1}-f_{i}=k_{i}zf_{i+1}}$ where ${\displaystyle k_{1}={\tfrac {a-b}{b(b+1)}},k_{2}={\tfrac {a+1}{(b+1)(b+2)}},k_{3}={\tfrac {a-b-1}{(b+2)(b+3)}},k_{4}={\tfrac {a+2}{(b+3)(b+4)}}}$, producing

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

or

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

Similarly

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

or

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

Since ${\displaystyle {}_{1}F_{1}(0;b;z)=1}$, setting a to 0 and replacing b + 1 with b in the first continued fraction gives a simplified special case:

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

The series ₂F₁

The final case involves

${\displaystyle {}_{2}F_{1}(a,b;c;z)=1+{\frac {ab}{c\,1!}}z+{\frac {a(a+1)b(b+1)}{c(c+1)\,2!}}z^{2}+{\frac {a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)\,3!}}z^{3}+\cdots .\,}$

Again, two identities are used alternately.

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

${\displaystyle \,_{2}F_{1}(a,b;c-1;z)-\,_{2}F_{1}(a,b+1;c;z)={\frac {(b-c+1)az}{c(c-1)}}\,_{2}F_{1}(a+1,b+1;c+1;z).}$

These are essentially the same identity with a and b interchanged.

Let

${\displaystyle f_{0}(z)=\,_{2}F_{1}(a,b;c;z),}$

${\displaystyle f_{1}(z)=\,_{2}F_{1}(a+1,b;c+1;z),}$

${\displaystyle f_{2}(z)=\,_{2}F_{1}(a+1,b+1;c+2;z),}$

${\displaystyle f_{3}(z)=\,_{2}F_{1}(a+2,b+1;c+3;z),}$

${\displaystyle f_{4}(z)=\,_{2}F_{1}(a+2,b+2;c+4;z),}$

etc.

This gives ${\displaystyle f_{i-1}-f_{i}=k_{i}zf_{i+1}}$ where ${\displaystyle k_{1}={\tfrac {(a-c)b}{c(c+1)}},k_{2}={\tfrac {(b-c-1)(a+1)}{(c+1)(c+2)}},k_{3}={\tfrac {(a-c-1)(b+1)}{(c+2)(c+3)}},k_{4}={\tfrac {(b-c-2)(a+2)}{(c+3)(c+4)}}}$, producing

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

or

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

Since ${\displaystyle {}_{2}F_{1}(0,b;c;z)=1}$, setting a to 0 and replacing c + 1 with c gives a simplified special case of the continued fraction:

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

The series ₀F₁

We have

${\displaystyle \cosh(z)=\,_{0}F_{1}({\tfrac {1}{2}};{\tfrac {z^{2}}{4}}),}$

${\displaystyle \sinh(z)=z\,_{0}F_{1}({\tfrac {3}{2}};{\tfrac {z^{2}}{4}}),}$

so

${\displaystyle \tanh(z)={\frac {z\,_{0}F_{1}({\tfrac {3}{2}};{\tfrac {z^{2}}{4}})}{\,_{0}F_{1}({\tfrac {1}{2}};{\tfrac {z^{2}}{4}})}}={\cfrac {z/2}{{\tfrac {1}{2}}+{\cfrac {\tfrac {z^{2}}{4}}{{\tfrac {3}{2}}+{\cfrac {\tfrac {z^{2}}{4}}{{\tfrac {5}{2}}+{\cfrac {\tfrac {z^{2}}{4}}{{\tfrac {7}{2}}+{}\ddots }}}}}}}}={\cfrac {z}{1+{\cfrac {z^{2}}{3+{\cfrac {z^{2}}{5+{\cfrac {z^{2}}{7+{}\ddots }}}}}}}}.}$

This particular expansion is known as Lambert's continued fraction and dates back to 1768.^[8]

It easily follows that

${\displaystyle \tan(z)={\cfrac {z}{1-{\cfrac {z^{2}}{3-{\cfrac {z^{2}}{5-{\cfrac {z^{2}}{7-{}\ddots }}}}}}}}.}$

The expansion of tanh can be used to prove that eⁿ is irrational for every non-zero integer n (which is alas not enough to prove that e is transcendental). The expansion of tan was used by both Lambert and Legendre to prove that π is irrational.

The Bessel function ${\displaystyle J_{\nu }}$ can be written

${\displaystyle J_{\nu }(z)={\frac {({\tfrac {1}{2}}z)^{\nu }}{\Gamma (\nu +1)}}\,_{0}F_{1}(\nu +1;-{\frac {z^{2}}{4}}),}$

from which it follows

${\displaystyle {\frac {J_{\nu }(z)}{J_{\nu -1}(z)}}={\cfrac {z}{2\nu -{\cfrac {z^{2}}{2(\nu +1)-{\cfrac {z^{2}}{2(\nu +2)-{\cfrac {z^{2}}{2(\nu +3)-{}\ddots }}}}}}}}.}$

These formulas are also valid for every complex z.

The series ₁F₁

Since ${\displaystyle e^{z}={}_{1}F_{1}(1;1;z)}$, ${\displaystyle 1/e^{z}=e^{-z}}$

${\displaystyle e^{z}={\cfrac {1}{1+{\cfrac {-z}{1+{\cfrac {z}{2+{\cfrac {-z}{3+{\cfrac {2z}{4+{\cfrac {-2z}{5+{}\ddots }}}}}}}}}}}}}$

${\displaystyle e^{z}=1+{\cfrac {z}{1+{\cfrac {-z}{2+{\cfrac {z}{3+{\cfrac {-2z}{4+{\cfrac {2z}{5+{}\ddots }}}}}}}}}}.}$

With some manipulation, this can be used to prove the simple continued fraction representation of e,

${\displaystyle e=2+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{4+{}\ddots }}}}}}}}}}}$

The error function erf (z), given by

${\displaystyle \operatorname {erf} (z)={\frac {2}{\sqrt {\pi }}}\int _{0}^{z}e^{-t^{2}}\,dt,}$

can also be computed in terms of Kummer's hypergeometric function:

${\displaystyle \operatorname {erf} (z)={\frac {2z}{\sqrt {\pi }}}e^{-z^{2}}\,_{1}F_{1}(1;{\scriptstyle {\frac {3}{2}}};z^{2}).}$

By applying the continued fraction of Gauss, a useful expansion valid for every complex number z can be obtained:^[9]

${\displaystyle {\frac {\sqrt {\pi }}{2}}e^{z^{2}}\operatorname {erf} (z)={\cfrac {z}{1-{\cfrac {z^{2}}{{\frac {3}{2}}+{\cfrac {z^{2}}{{\frac {5}{2}}-{\cfrac {{\frac {3}{2}}z^{2}}{{\frac {7}{2}}+{\cfrac {2z^{2}}{{\frac {9}{2}}-{\cfrac {{\frac {5}{2}}z^{2}}{{\frac {11}{2}}+{\cfrac {3z^{2}}{{\frac {13}{2}}-{\cfrac {{\frac {7}{2}}z^{2}}{{\frac {15}{2}}+-\ddots }}}}}}}}}}}}}}}}.}$

A similar argument can be made to derive continued fraction expansions for the Fresnel integrals, for the Dawson function, and for the incomplete gamma function. A simpler version of the argument yields two useful continued fraction expansions of the exponential function.^[10]

The series ₂F₁

From

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

${\displaystyle (1-z)^{-b}={\cfrac {1}{1+{\cfrac {-bz}{1+{\cfrac {(b-1)z}{2+{\cfrac {-(b+1)z}{3+{\cfrac {2(b-2)z}{4+{}\ddots }}}}}}}}}}}$

It is easily shown that the Taylor series expansion of arctan z in a neighborhood of zero is given by

${\displaystyle \arctan z=zF({\scriptstyle {\frac {1}{2}}},1;{\scriptstyle {\frac {3}{2}}};-z^{2}).}$

The continued fraction of Gauss can be applied to this identity, yielding the expansion

${\displaystyle \arctan z={\cfrac {z}{1+{\cfrac {(1z)^{2}}{3+{\cfrac {(2z)^{2}}{5+{\cfrac {(3z)^{2}}{7+{\cfrac {(4z)^{2}}{9+\ddots }}}}}}}}}},}$

which converges to the principal branch of the inverse tangent function on the cut complex plane, with the cut extending along the imaginary axis from i to the point at infinity, and from −i to the point at infinity.^[11]

This particular continued fraction converges fairly quickly when z = 1, giving the value π/4 to seven decimal places by the ninth convergent. The corresponding series

${\displaystyle {\frac {\pi }{4}}={\cfrac {1}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+\ddots }}}}}}}}=1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}\pm \cdots }$

converges much more slowly, with more than a million terms needed to yield seven decimal places of accuracy.^[12]

Variations of this argument can be used to produce continued fraction expansions for the natural logarithm, the arcsin function, and the generalized binomial series.