Jtdylktuy

I was looking at this paper and saw this nice sum in section [12] of the paper,

$$sum_{n=0}^{infty}frac{1}{4n+1}left[frac{1}{4^n}{2n choose n}right]^2=frac{Gamma^4left(frac{1}{4}right)}{16pi^2}tag1$$

out of curiosity I conjectured the following two sums

$$begin{align*}
sum_{n=0}^{infty}frac{(-1)^n}{(2n-1)^2}left[frac{1}{4^n}{2n choose n}right]^2&=frac{4sqrt{2pi}}{Gamma^2left(frac{1}{4}right)}tag2\
sum_{n=0}^{infty}(-1)^nfrac{n^2}{(2n-1)^2}left[frac{1}{4^n}{2n choose n}right]^2&=-frac{Gamma^2left(frac{1}{4}right)}{8pisqrt{2pi}}tag3
end{align*}$$

I am unable to prove them.

How do we go about to prove these two sums?

From the beautiful answer by @robjohn, we learn
begin{equation}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{(2k-1)^2}
=frac2{pi r}int_0^1frac{left(rxarcsin(rx)+sqrt{1-r^2x^2}right),mathrm{d}x}{sqrt{1-x^2}}
end{equation}
Plugging in $r=i$,
begin{align}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{(-1)^k}{(2k-1)^2}
&=frac{2}{pi}int_0^1frac{left(ixarcsin(ix)+sqrt{1+x^2}right),mathrm{d}x}{sqrt{1-x^2}}\
&=frac{2}{pi}int_0^1frac{sqrt{1+x^2}-xsinh^{-1}(x)}{sqrt{1-x^2}},mathrm{d}x\
&=frac{2}{pi}int_0^1left[ frac{sqrt{1+x^2}}{sqrt{1-x^2}}-frac{sqrt{1-x^2}}{sqrt{1+x^2}}right],mathrm{d}x\
&=frac{4}{pi}int_0^1 frac{x^2}{sqrt{1-x^4}},mathrm{d}x\
&=frac{1}{pi}Bleft( frac{3}{4},frac{1}{2} right)\
&=frac{4sqrt{2pi}}{Gamma^2left( frac{1}{4} right)}
end{align}
(the integration of the $sinh^{-1}$ term was made by parts).

For the second expression, we decompose
begin{equation}
frac{n^2}{left( 2n-1 right)^2}=frac{1}{4}+frac{1}{2}frac{1}{2n-1}+frac{1}{4}frac{1}{left( 2n-1 right)^2}
end{equation}
From the linked answer we have also
begin{equation}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{2k-1}
=frac1{pi r}int_0^1frac{-sqrt{1-r^2x},mathrm{d}x}{sqrt{x(1-x)}}
end{equation}
and thus, with $r=i$
begin{align}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{(-1)^k}{2k-1}
&=-frac1{pi }int_0^1frac{sqrt{1+x},mathrm{d}x}{sqrt{x(1-x)}}\
&=-frac2{pi }int_0^1frac{sqrt{1+t^2}}{sqrt{1-t^2}},mathrm{d}t\
&=-frac{2sqrt{2}}{pi}Eleft( frac{1}{sqrt{2}} right)\
&=-frac{2sqrt{2}}{pi}left[ frac{Gamma^2left( frac{1}{4} right)}{8sqrt{pi}}+frac{pi^{3/2}}{Gamma^2left( frac{1}{4} right)}right]
end{align}
where the singular value $k_1$ for the elliptic integral is used. We may also express from the answer
begin{equation}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}r^{2k}
=frac1piint_0^1frac{mathrm{d}x}{sqrt{1-r^2x}sqrt{x(1-x)}}
end{equation}
which, with $r=i$ again, gives
begin{align}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}(-1)^{k}
&=frac1piint_0^1frac{mathrm{d}x}{sqrt{1+x}sqrt{x(1-x)}}\
&=frac2piint_0^1frac{dt}{sqrt{1-t^4}}\
&=frac{1}{2pi}Bleft(frac{1}{4}, frac{1}{2} right)\
&=frac{sqrt{2pi}}{2Gamma^2left(frac{3}{4}right)}
end{align}
Then,
begin{align}
S&=sum_{n=0}^{infty}frac{(-1)^nn^2}{(2n-1)^2}left[frac{1}{4^n}{2n choose n}right]^2\
&=frac{sqrt{2pi}}{8Gamma^2left(frac{3}{4}right)}
-frac{sqrt{2}}{pi}left[ frac{Gamma^2left( frac{1}{4} right)}{8sqrt{pi}}+frac{pi^{3/2}}{Gamma^2left( frac{1}{4} right)}right]+frac{sqrt{2pi}}{Gamma^2left( frac{1}{4} right)}
end{align}
Using the reflection formula $Gamma(3/4)Gamma(1/4)=pisqrt{2}$, we obtain finally
begin{equation}
S=-frac{Gamma^2left( frac{1}{4} right)}{8sqrt2pi^{3/2}}
end{equation}
as expected.

Edit (simpler derivation for the second expression)

For the second expression, it is easier to leave the integrals unevaluated in the decomposition:
begin{equation}
S=int_0^1left[frac{1}{2pi}frac{1}{sqrt{1-x^4}}-frac{1}{pi}sqrt{frac{1+x^2}{1-x^2}}+frac{1}{pi}frac{x^2}{sqrt{1-x^4}} right],dx
end{equation}
or
begin{align}
S&=-frac{1}{2pi}int_0^1frac{dx}{sqrt{1-x^4}}\
&=-frac{1}{8pi}int_0^1 left( 1-u right)^{-1/2}u^{-3/4},du\
&=-frac{Bleft( frac{1}{4},frac{1}{2} right)}{8pi}
end{align}
and the result follows from the $Gamma$ reflection formula.