Convergence of Hypergeometric pFq

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Consider the series ${}_pF_q(a_1,a_2,\ldots,a_p;b_1,b_2,\ldots,b_q;z)$, where ${}pF_q$ denotes hypergeometric pFq. Then,

  • If any of the $a_j$'s is a a nonpositive integer, then the series terminates and is a polynomial.
  • If any of the $b_{\ell}$'s is a nonpositive integer, the series diverges because of divison by zero.
  • The remaining convergence of the series can be split into three cases:
    • $p<q+1$
    • $p=q+1$
    • $p>q+1$

Case I: $p<q+1$


The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.


If $t=0$ then the series converges trivially, so suppose $t \neq 0$. We will apply the ratio test. Let $\alpha_k=\dfrac{\vec{a}^{\overline{k}}t^k}{\vec{b}^{\overline{k}}k!}$. Then $$\begin{array}{ll} L &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\alpha_{k+1}}{\alpha_k} \right| \\ &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\dfrac{\vec{a}^{\overline{k}}t^k}{\vec{b}^{\overline{k}}k!}}{\dfrac{\vec{a}^{\overline{k+1}}t^{k+1}}{\vec{b}^{\overline{k+1}}(k+1)!}} \right| \\ &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{\vec{a}^{\overline{k}} \vec{b}^{\overline{k+1}}(k+1)!t^k }{\vec{b}^{\overline{k}} \vec{a}^{\overline{k+1}}k!t^{k+1}} \right| \\ &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{k(\vec{b}+k)}{(\vec{a}+k)t} \right| \\ &= \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{O(k^{q+1})}{O(k^{p})}\right| \\ &= 0 < 1, \end{array}$$ therefore the series converges for all $t \in \mathbb{C}$. █ </div></div>

Case II: $p=q+1$

Proposition: The hypergeometric pFq ${}_pF_q$ converges for all $t\in \mathbb{C}$ with $|t|<1$.

Proof: █

Case III: $p>q+1$

Proposition: The series ${}_pF_q$ diverges for all $t \in \mathbb{C}$.

Proof: █