Is there an automated way to express hypergeometric functions in series form using gamma functions, factorials, double factorials or rising factorials? For example using the formula (on the hypergeometric function wiki page):

$$_2 F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}$$

or similar expressions for the general $_p F_q$ case. I work with the series/sum notation and occasionally Mathematica gives hypergeometric expressions which can be hard to read and annoying to convert manually.

  • 1
    $\begingroup$ Sometimes FunctionExpand will convert hypergeometric functions to simpler form, but don't count on it working too often. You have a better chance with numbers than with variables, and with variables you have a better chance with Assumptions. Asking for the Series representation may work, but probably they will not be in terms of Gamma functions. $\endgroup$
    – Bill Watts
    Jul 15, 2019 at 7:43
  • $\begingroup$ You can go to Hypergeometric2F1 definition and copy the InputForm version of the definition and paste it into your notebook, replacing Sum with Inactive[Sum] to keep it in unevaluated form. Then you can replace individual values and later use Activate on the expression to resume evaluation if you want. $\endgroup$ Jul 15, 2019 at 16:46

2 Answers 2


Try the following code:

HypergeometricPFQk[a_List, b_List, z_, k_] := z^k/k! Product[Pochhammer[ak, k], {ak, a}] / Product[Pochhammer[bk, k], {bk, b}];

For example of use:

HypergeometricPFQk[{a, b}, {c}, z, k] // TeXForm

$$ \frac{z^k (a)_k (b)_k}{k! (c)_k} $$

Alternatively use:

HypergeometricPFQk[{a, b}, {c}, z, k] // FunctionExpand // InputForm

which returns

(z^k*Gamma[c]*Gamma[a + k]*Gamma[b + k])/(Gamma[a]*Gamma[b]*Gamma[1 + k]*Gamma[c + k])

or in TeXForm

$$ \frac{\Gamma (c) z^k \Gamma (a+k) \Gamma (b+k)}{\Gamma (a) \Gamma (b) \Gamma (k+1) \Gamma (c+k)} $$


One possible workaround: Assuming[n >= 0, SeriesCoefficient[HyperGeometricPFQ[{a1,a2,a3...},{b1,b2,b3...}, z],{z,0,n}]] gives an expression in terms of factorials. If $z$ is fixed in the original hypergeometric function, substitute a variable $z$ so the series has a variable to work with.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.