Sum of squared hypergeometric polynomials

Question

$\sum_{m=1}^\infty \frac{1}{m} \bigg[{}_2F_1(-m,m,2,u)\bigg]^2 = \frac 1 4 -\frac 1 2 \log u$

I have very strong numerical support that this is true when $0<u\le1$. Can anyone help proving or disproving ?

Answer 1

The identity can be derived by differentiating both sides with respect to $u$:

The rhs \begin{align}\tag{1}\label{eq:1} R(u)=\frac 1 4-\frac 1 2 \log u \end{align} simply gives \begin{align}\tag{2}\label{eq:2} R'(u)=-\frac{1}{2u}. \end{align} The lhs $L(u)$ is first written as hypergeometric double series, \begin{align}\tag{3}\label{eq:3} L(u) = \sum_{m=1}^\infty \frac 1 m \sum_{k=0}^\infty \sum_{l=0}^\infty \frac{ (-m)_k (m)_k \,(-m)_l (m)_l}{ (2)_k \, (2)_l} \frac{u^{k+l}}{k! \, l!} , \end{align} such that \begin{align}\tag{4}\label{eq:4} L'(u) = \sum_{m=1}^\infty \frac 1 m \sum_{k=0}^\infty \sum_{l=0}^\infty (k+l) \frac{(-m)_k (m)_k (-m)_l (m)_l}{(2)_k \,(2)_l} \frac{u^{k+l-1}}{k! \, l!}. \end{align} At this point we can reorder the sums and first evaluate the sum over $m$, \begin{align}\tag{5}\label{eq:5} L'(u) &= \sum_{k=0}^\infty \sum_{l=0}^\infty \sum_{m=1}^\infty \frac {k+l} m \frac{(-m)_k (m)_k (-m)_l (m)_l}{(2)_k \,(2)_l} \frac{u^{k+l-1}}{k! \, l!}\\ \tag{6}\label{eq:6} &= -\sum_{k=0}^\infty \sum_{l=0}^\infty \frac{(k-1) (l-1) (-1)_k (-1)_l}{2 (2)_k \,(2)_l} u^{k+l-1}. \end{align} We are lucky now, as the only term contributing to this double sum is the term with $k=l=0$, such that \begin{align}\tag{7}\label{eq:7} L'(u) &= -\frac{1}{2u}. \end{align} Note that this simple result was not obtained without the initial differentiation, as then the $k{=}l{=}0$ term diverges.

The remaining step is the determination of the integration constant, which can most easily evaluated at $u=1$, where the sum trivially gives $L(1)=R(1)=1/4$. Note that here only the first term $m=1$ contributes.

Answer 2

I think the proof below should work. Although it looks more complicated than the one of Fred Hucht, it doesn't involve divergent series.

I write \begin{equation}\tag{1}S(t)=\sum_{m=1}^\infty\frac {t^m}m\, {}_2F_1\left(\begin{matrix}-m,m\\2\end{matrix};u\right)^2, \end{equation} where $t^m$ is inserted to improve convergence. We are interested in the value $S=S(1)$.

The key fact that I use is Bateman's product formula \begin{multline*}{}_2F_1\left(\begin{matrix}-m,a+m\\b\end{matrix};u\right)\,{}_2F_1\left(\begin{matrix}-m,a+m\\b\end{matrix};v\right)\\ =(-1)^m\frac{(1+a-b)_m}{(b)_m}\sum_{k=0}^m \frac{(-m)_k(a+m)_k}{k!(1+a-b)_k}(1-u-v)^k\,{}_2F_1\left(\begin{matrix}-k,a+k\\b\end{matrix};-\frac{uv}{1-u-v}\right). \end{multline*} In the case of interest, $a=0$ and $b=2$, we need to interpret $$\frac{(1+a-b)_m}{(1+a-b)_k}=(1+a-b+k)_{m-k}=(k-1)_{m-k}, $$ which vanishes if $m\geq 2$ and $0\leq k\leq 1$. That is, for $m\geq 2$, \begin{multline}\label{b}{}_2F_1\left(\begin{matrix}-m,m\\2\end{matrix};u\right){}_2F_1\left(\begin{matrix}-m,m\\2\end{matrix};v\right)\\ =\frac{(-1)^m}{(2)_m}\sum_{k=2}^m \frac{(-m)_k(m)_k(k-1)_{m-k}}{k!}(1-u-v)^k\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{uv}{1-u-v}\right). \end{multline}

In $S(t)$, we isolate the term with $m=1$. In the remaining terms, insert (1), change the order of summation and replace $m$ by $m+k$. After simplification, this gives \begin{multline*}S(t)=t\left(1-\frac u2\right)^2 + \sum_{k=2}^\infty\frac{(2k-1)!}{k!(k+1)!}\,t^k (1-2u)^k\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{u^2}{1-2u}\right)\\ \times \sum_{m=0}^\infty\frac{(2k)_m(k-1)_m}{m!(k+2)_m}\,(-t)^m.\end{multline*} Note that if we formally put $t=1$ and use Kummer's summation formula $${}_2F_1\left(\begin{matrix}a,b\\1+a-b\end{matrix};-1\right)=\frac{\Gamma(1+a-b)\Gamma(1+a/2)}{\Gamma(1+a)\Gamma(1+a/2-b)}, $$ the sum in $m$ evaluates to $(k+1)!k!/(2k)!$. However, for $t=1$ this sum is divergent. To remedy this, we first apply the linear transformation $${}_2F_1\left(\begin{matrix}2k,k-1\\k+2\end{matrix};-t\right)=(1+t)^{3-2k}\,{}_2F_1\left(\begin{matrix}2-k,3\\k+2\end{matrix};-t\right). $$ We can then apply the terminating case of Kummer's formula, which gives the same result as was obtained formally: $$\lim_{t\rightarrow 1}{}_2F_1\left(\begin{matrix}2k,k-1\\k+2\end{matrix};-t\right)=2^{3-2k}\,{}_2F_1\left(\begin{matrix}2-k,3\\k+2\end{matrix};-1\right) =2^{3-2k}\frac{(4)_{k-2}}{(5/2)_{k-2}}=\frac{(k+1)!k!}{(2k)!}. $$ Assuming that we can interchange limit and summation (something needs to be checked here), it follows that $$S=\left(1-\frac u2\right)^2+ \frac 12\sum_{k=2}^\infty\frac{(1-2u)^k}{k}\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{u^2}{1-2u}\right). $$ If we start the summation at $k=1$, we add the term $(1-u/2)^2-1/2$. Thus, $$S =\frac 12+\frac 12\sum_{k=1}^\infty\frac{(1-2u)^k}{k}\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{u^2}{1-2u}\right). $$ We write the final ${}_2F_1$ as a sum over $j$. We split of the term corresponding to $j=0$, which gives the contribution $$\frac 12\sum_{k=1}^\infty\frac{(1-2u)^k}{k}=-\frac 12\,\log(2u).$$ In the remaining terms, we replace $k$ by $k+j$ and change the order of summation. This gives \begin{multline*}\frac 12\sum_{k=1}^\infty\frac{(1-2u)^k}{k}\sum_{j=1}^{k}\frac{(-k)_j(k)_j}{j!(2)_j}\left(-\frac{u^2}{1-2u}\right)^j \\ =\frac 12\sum_{j=1}^\infty\frac{(2j-1)!}{j!(j+1)!}\,u^{2j}\sum_{k=0}^\infty\frac{(2j)_k}{k!}(1-2u)^k =\frac 12\sum_{j=1}^\infty\frac{(2j-1)!}{j!(j+1)!}\frac 1{4^j} =-\frac 14+\frac{\log 2}2, \end{multline*} where we used the non-terminating binomial theorem to compute the sum over $k$. The final step can be obtained from the Taylor expansion $$\sum_{j=1}^\infty\frac{(2j-1)!}{j!(j+1)!}\,x^j=\log(2)-\log(1+\sqrt{1-4x})+\frac{\sqrt{1-4x}+2x-1}{4x}.$$ Collecting the different contributions finally gives $$S=\frac 12-\frac 12\,\log(2u)-\frac 14+\frac{\log 2}2=\frac14-\frac12\log(u). $$

Answer 3

The identity can also be shown without divergences using generating functions, as suggested in the comment of @TimothyBudd above.

Using the generating functions \begin{align} \tag{1a}\label{eq:1a} f(u,t)&=\frac{t-t^{-1}}{4u}\left( 1-\frac{1}{1-t}\sqrt{(1-t)^2+4 u t}\right) -\frac 1 2\\ \tag{1b}\label{eq:1b} &=\sum_{m=1}^\infty {}_{2}F_1(-m,m;2;u)\,t^m \end{align} as well as \begin{align} \tag{2a}\label{eq:2a} g(u,t)&=\int_0^{t} \mathrm d t'\frac{f(u,t')}{t'} \\ \tag{2a}\label{eq:2b} &=\frac{t-1}{t+1}\left(f(u,t)+\frac 1 2\right) +\log\left(2 \,\frac{f(u,t)+\frac 1 2}{t+1}\right) + \frac 1 2\\ \tag{2c}\label{eq:2c} &=\sum_{m=1}^\infty \frac{\,{}_{2}F_1(-m,m;2;u)}{m}\, t^m \,, \end{align} we can apply the Hadamard product formula at $t=1$ to get \begin{align} \tag{3a}\label{eq:3a} L(u)&= \sum_{m=1}^\infty \frac {\,{}_{2}F_1(-m,m;2;u)^2}{m} \\ \tag{3b}\label{eq:3b} &= \frac{1}{2\pi i} \oint_{|z|=1} \mathrm d z \,z^{-1}f(u,z^{\pm 1})\,g(u,z^{\mp 1}) \,. \end{align} The analytic evaluation of this unit circle contour integral is left as an exercise. Numerically it gives the desired result.