Showing that $ { }_2F_1(1, n + 1;n+2; \frac{1}{2}) \in O(2^n)$

Question

Is the following statement true? $$ { }_2F_1\left(1, n + 1;n+2; \frac{1}{2}\right) \in O(2^n)$$

What are the steps to prove it?

Answer 1

By way of "enrichment" we can obtain the formula for the hypergeometric value using the Barnes integral. According to Wikipedia we have the following integral: $$_2F_1(a,b;c; z) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \times \frac{1}{2\pi i} \int_{-i\infty}^{+i\infty} \frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)} (-z)^s ds.$$ Subsituting the values for $a,b$ and $c$ we get $$_2F_1(1,n+1;n+2; 1/2) \\= \frac{\Gamma(n+2)}{\Gamma(1)\Gamma(n+1)} \times \frac{1}{2\pi i} \int_{-i\infty}^{+i\infty} \frac{\Gamma(1+s)\Gamma(n+1+s)\Gamma(-s)}{\Gamma(n+2+s)} (-1/2)^s ds \\ = - (n+1)\times \frac{1}{2\pi i} \int_{-i\infty}^{+i\infty} \frac{\pi}{\sin(\pi s)} \times \frac{1}{1+n+s} (1/2)^s e^{\pi i s} ds.$$ Evaluating this in the right half plane with the Cauchy Residue Theorem we get $$(n+1)\sum_{q\ge 0} (-1)^q \times \frac{1}{n+1+q} (1/2)^q e^{i\pi q}. = (n+1) \sum_{q\ge 0} \frac{1}{n+1+q} (1/2)^q.$$ This gives the asymptotic expansion $$\sum_{q\ge 0} \frac{1}{1+q/(n+1)} (1/2)^q \\ \sim \sum_{q\ge 0} (1/2)^q - \frac{1}{n+1}\sum_{q\ge 0} q(1/2)^q + \frac{1}{(n+1)^2}\sum_{q\ge 0} q^2(1/2)^q - \frac{1}{(n+1)^3}\sum_{q\ge 0} q^3(1/2)^q +\cdots$$ which is $$2 - \frac{2}{n+1} + \frac{6}{(n+1)^2} - \frac{26}{(n+1)^3} +\cdots$$ confirming the result from the definitive answer.

Remark. The above is definitely an asymptotic expansion and not a convergent series. Care must be taken when using it for calculations. The rule is to take as many initial terms as possible before they start diverging. E.g. for $n=25$ the first nine / ten terms give $$1.930731710\quad\text{and}\quad 1.930733159$$ whereas the exact value is $$1.930732621 .$$

Digression. The reason that this is an asymptotic expansion as opposed to a convergent series lies in the fact that $$\sum_{q\ge 0} q^n (1/2)^q = \mathrm{Li}_{-n}(1/2) \sim \frac{n!}{(\log 2)^{n+1}},$$ so that the coefficients outgrow any polynomial in $n$ as documented for example at the OEIS entry A076726. (In what follows the $n$ is not the same as in the first part of this post.)

We now prove this as it is a calculation worth knowing. Start with $$\mathrm{Li}_{-n}(z) = \sum_{k=0}^k k! {n+1 \brace k+1} \left(\frac{z}{1-z}\right)^{k+1}$$ and note that for $z=1/2$ the rational power disappears so that $$\mathrm{Li}_{-n}(1/2) = \sum_{k=0}^n k! {n+1 \brace k+1}.$$ Now the bivariate mixed generating function of the Stirling numbers of the second kind is $$\exp(u(\exp(z)-1))$$ so that $${n+1\brace k+1} = (n+1)! [z^{n+1}] [u^{k+1}] \exp(u(\exp(z)-1)) = (n+1)! [z^{n+1}] \frac{(\exp(z)-1)^{k+1}}{(k+1)!}.$$ Returning to the sum we get $$(n+1)! [z^{n+1}] \sum_{k=0}^n \frac{(\exp(z)-1)^{k+1}}{k+1}.$$ We may extend the sum to infinity as the extra terms do not contribute to $[z^{n+1}]$ to get $$(n+1)! [z^{n+1}] \sum_{k=0}^\infty \frac{(\exp(z)-1)^{k+1}}{k+1} \\= (n+1)! [z^{n+1}] \log \frac{1}{1-(\exp(z)-1)} = (n+1)! [z^{n+1}] \log \frac{1}{2-\exp(z)} = (n+1)! [z^{n+1}] \left(\log \frac{1}{2} + \log \frac{1}{1-\exp(z)/2}\right).$$ Therefore we seek the asymptotics of the following term: $$(n+1)! [z^{n+1}] \log \frac{1}{1-\exp(z)/2}.$$ This has branch points at $\rho_m = \log 2 + 2\pi i m,$ with $m$ an integer. The one that is closest to the origin is $\rho_0 = \log 2,$ so we seek an expansion about this point.

We observe that the pole of the inner term is simple as we compute the residue: $$\mathrm{Res}\left(\frac{1}{1-\exp(z)/2}; z=\log 2\right) =\left. \frac{1}{-\exp(z)/2}\right|_{z=\log 2} = -1.$$ Now considering the Laurent expansion of the inner term the simple pole dominates in a neighborhood of $\log 2$ while the higher order terms vanish, giving $$\log \left(-\frac{1}{z-\log 2}\right) = \log\left(\frac{1}{\log 2 - z}\right) = \log\frac{1}{\log 2}+ \log\left(\frac{1}{1-z/\log 2}\right).$$ We therefore get the following asymptotics $$(n+1)! [z^{n+1}] \log\left(\frac{1}{1-z/\log 2}\right) = (n+1)! \frac{(1/\log 2)^{n+1}}{n+1} = \frac{n!}{(\log 2)^{n+1}}$$ as was to be shown.

Replacing $n$ by $q$ and returning to the initial computation we get for the asymptotics of the term with index $q$ the formula $$(-1)^q \frac{q!}{(\log 2)^{q+1}\times (n+1)^q}$$ which is now clearly seen to eventually diverge.

Answer 2

Simply plugging the arguments into the definition of the hypergeometric function shows that $$ {}_2F_1\big( 1,n+1;n+2;\tfrac12 \big) = \sum_{j=0}^\infty \frac{n+1}{2^j(n+1+j)} < \sum_{j=0}^\infty \frac1{2^j} = 2. $$ In fact it is an increasing function of $n$ that tends to $2$ from below as $n$ tends to infinity.

So your statement is true (assuming you mean for $n$ to stay rather larger than $-2$) but much weaker than necessary.