Difference between revisions of "Convergence of Hypergeometric pFq"
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==Case II: $p=q+1$== | ==Case II: $p=q+1$== | ||
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| − | <strong>Proposition:</strong> The | + | <strong>Proposition:</strong> The [[hypergeometric pFq]] ${}_pF_q$ converges for all $t\in \mathbb{C}$ with $|t|<1$.<br /> |
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<strong>Proof: █</strong> | <strong>Proof: █</strong> | ||
Revision as of 21:28, 26 June 2016
Convergence
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:
Case I: $p<q+1$
Proposition: The series ${}_pF_q$ converges for all $t \in \mathbb{C}$.
Proof: Notice 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 $t \in \mathbb{C}$. █
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: █
