Convergence or divergence of the gaussian hypergeometric function $_{2}F_1(a,b;c;z)$ where $a,b$ and $c$ are complex.

Question

I'm studying the convergence of the hypergeometric function and I have some questions. First, I define the hypergeometrical functions as \begin{equation} _{2}F_{1} (a,b;c;z)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(a)_n(b)_n}{(c)_n}\dfrac{1}{n!}z^n, \end{equation} where $(w)_n$ is the rising factorial and it is defined by \begin{equation} (w)_n=\left\lbrace \begin{array}{cc} 1 & \text{for}~n=0\\ w(w+1)\cdots (w-n+1) & \text{for}~n\neq 0 \end{array}\right. \end{equation}
and the parameters $a$, $b$ and $c$ are complex.

Then, I know that:

• For $|z|=1,z\neq 1$, we have that it is not absolutely convergent, but convergent for $0\leq \Re(a+b-c)<1$. But I don't know how to prove that it is convergent there. Moreover, in this case, if $\Re(a+b-c)>1$ we obtain that it is divergent and if $\Re(a+b-c)=1$ and $\Re(a+b)\leq\Re(ab)$ it is also divergent. But I don't know again how to see it.

Thanks a lot.

Answer 1

Assume that $a$, $b$, and $c$ are complex parameters.

For large $n$, we have $$ \begin{align}\frac{(a)_n(b)_n}{(c)_n}\frac{1}{n!} &= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} \frac{\Gamma(a+n) \Gamma(b+n)}{\Gamma(c+n)} \frac{1}{\Gamma(1+n)} \frac{\Gamma(n)}{\Gamma(n)} \\ &= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} \frac{\Gamma(a+n)}{\Gamma(n)} \frac{\Gamma(b+n)}{\Gamma(n)} \frac{\Gamma(n)}{\Gamma(c+n)} \frac{1}{n} \\ & \sim \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} n^{a}n^{b}n^{-c}n^{-1} \tag{1} \\&= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} n^{a+b-c-1}. \end{align} $$

Therefore, by the p-series test, the series converges absolutely for $|z|=1$ and $\Re(a + b -c) <0$.

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$(1)$ $\Gamma(\beta+n) \sim \Gamma(n)n^{\beta} $ as $n \to \infty$

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To show that the series converges conditionally for $|z|=1$, $z \ne 1$, and $0 \le \Re(a+b-c) <1$, we can use the proof of Dirichlet's test .

Saying that $|z|=1$, $z \ne1$, is equivalent to saying that $z^n= e^{i n \theta} = \cos(n \theta) + i \sin(n\theta)$ for some $\theta$ between $0$ and $2 \pi$.

And the partial sums of $\cos(n \theta)$ and $\sin (n \theta)$ are bounded. (See here, for example.)

Looking at the proof of Dirichlet's test, we need to show that $$ \sum_{n=1}^{\infty} \left| n^{a+b+c-1} - (n+1)^{a+b+c-1}\right| \tag{2}$$ converges for $\Re(a+b-c) < 1$.

To show that it converges, let $\alpha = a+b-c$.

Then for large $n$, we have $$ \begin{align} n^{\alpha-1}-(n+1)^{\alpha-1} &= n^{\alpha-1}-n^{\alpha-1} \left(1+ \frac{1}{n} \right)^{\alpha-1} \\ &=n^{\alpha-1} -n^{\alpha-1} \left(1+ \frac{\alpha-1}{n} + \mathcal{O}(n^{-2}) \right) \\ &= (1- \alpha) n^{\alpha-2} + \mathcal{O}(n^{\alpha-3}). \end{align}$$

Therefore, as $n \to \infty$, $$|n^{\alpha-1} - (n+1)^{\alpha-1}| \sim |1-\alpha|n^{\Re(\alpha)-2},$$ which, by the p-series test, shows that $(2)$ converges for $\Re(\alpha) = \Re(a +b -c) < 1$.

Answer 2

Let us rewrite \begin{align} _2F_1(a, b, c; z):=\sum^\infty_{n=0}u_nz^n\tag{0}\label{zero} \end{align} where $u_n:=\frac{(a)_n(b)_n}{(c_n)n!}$, $a,b,c\in\mathbb{C}\setminus(\mathbb{Z}_-)$ where $\mathbb{Z}_-:=\{0\}\cup(-\mathbb{N})$, and $(z)_n:=\prod^{n-1}_{k=0}(z+k)$, $(z)_0:=1$.

Since \begin{align} \frac{u_{n+1}}{u_n}&=\frac{(a+n)(b+n)}{(c+n)(n+1)}=\frac{n^2+n(a+b)+ab}{n^2+n(c+1)+c}\\ &=1+\frac{n(a+b-c-1)+ab-c}{n^2+n(c+1) +c}\\ &=1+\frac{\xi}{n z_n}+o(1/n) \end{align} where $\xi=a+b-c-1\in\mathbb{C}$, $(z_n:n\in\mathbb{N})\subset \mathbb{C}$ and $z_n\xrightarrow{n\rightarrow\infty}1$, the radius of convergence of $_2F_1(a, b, c; z)$ is $1$ and so the series \eqref{zero} converges for all $|z|<1$.

Notice that \begin{align} n\Big(1-\frac{|u_{n+1}|}{|u_n|}\Big)&=n\Big(1-\Big|1+\frac{\xi}{n z_n}+o(1/n)\Big|\Big)\\ &=n\left(\frac{1-|1+\frac{\xi}{n z_n}+o(1/n)|^2}{1+|1+\frac{\xi}{n z_n}+o(1/n)|}\right)\\ &=n\left(\frac{-2\mathfrak{R}(\tfrac{\xi}{nz_n}+o(1/n))-\mathfrak{R}^2(\tfrac{\xi}{nz_n}+o(1/n))-\mathfrak{I}^2(\tfrac{\xi}{nz_n}+o(1/n))}{1+|1+\frac{\xi}{n z_n}+o(1/n)|}\right)\\ &=\frac{-2\mathfrak{R}(\frac{\xi}{z_n}+no(1/n))-\tfrac{1}{n}\mathfrak{R}^2(\tfrac{\xi}{z_n}+no(1/n))-\tfrac{1}{n}\mathfrak{I}^2(\frac{\xi}{z_n}+no(1/n))}{1+|1+\frac{\xi}{n z_n}+o(1/n)|} \end{align} Hence $$n\Big(1-\Big|\frac{u_{n+1}}{u_n}\Big|\Big)\xrightarrow{n\rightarrow\infty}-\mathfrak{R}(\xi)=1-\mathfrak{R}(a+b-c) $$

Now it is easy to decide absolute convergence of the series along the unit circle trough Raabe's test.