View source for Convergence of Hypergeometric pFq

__NOTOC__
==Theorem==
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: $p<q+1$==
===Proposition:===
The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.<br />
===Proof:===
If $t=0$ then the series converges trivially, so suppose $t \neq 0$. We will apply the [[ratio_test | 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 $t \in \mathbb{C}$. █
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==Case II: $p=q+1$==
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<strong>Proposition:</strong> The [[hypergeometric pFq]] ${}_pF_q$ converges for all $t\in \mathbb{C}$ with $|t|<1$.<br />
<div class="mw-collapsible-content">
<strong>Proof: █</strong>
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==Case III: $p>q+1$==
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
<strong>Proposition:</strong> The series ${}_pF_q$ diverges for all $t \in \mathbb{C}$.<br />
<div class="mw-collapsible-content">
<strong>Proof: █</strong>
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