Difference between revisions of "Convergence of Hypergeometric pFq"

Latest revision as of 14:32, 7 October 2016

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 $t \in \mathbb{C}$.

Proof:

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: $p=q+1$

Case III: $p>q+1$