For n
even, the Sum
in the question can be performed explicitly,
Evaluate[Unevaluated[n*Sum[Binomial[2 n - 4 i, n - 2 i]*Binomial[n, 2 i]*
Binomial[4 i, 2 i], {i, 0, Floor[n/2]}]/2^(3 n - 1)] /.
n -> 2 m /. Floor[(2*m)/2] -> m] // Simplify
(* 4^(1 - 3 m) m Binomial[4 m, 2 m] HypergeometricPFQ[
{1/4, 3/4, 1/2 - m, 1/2 - m, -m, -m}, {1/2, 1/2, 1, 1/4 - m, 3/4 - m}, 1] *)
and then partially expanded
f65 = Simplify[Series[%, {m, Infinity, 0}] // Normal, m > 0 && m ∈ Integers]
(* (2^(3/2 - 2 m) Sqrt[m] HypergeometricPFQ[{1/4, 3/4, 1/2 - m, 1/2 - m, -m, -m},
{1/2, 1/2, 1, 1/4 - m, 3/4 - m}, 1])/Sqrt[π] *}
(A similar calculation can, of course, be obtained for n
odd.) Next, f65
is represented as a series of Pochhammer
symbols.
2^(3/2 - 2 m) Sqrt[m/π] Sum[
Pochhammer[1/4, k] Pochhammer[3/4, k] Pochhammer[-m, k]^2 Pochhammer[1/2 - m, k]^2/
(Pochhammer[1/2, k]^2 Pochhammer[1, k] Pochhammer[1/4 - m, k] Pochhammer[3/4 - m, k])/k!
, {k, 0, ∞}]
Large m
expansions of the sort described in 32950 only work for ratios of equal numbers of Pochhammer
symbols involving m
. So, expand just the ratio of the last two Pochhammer
symbols in the numerator and denominator.
Series[Pochhammer[1/2 - m, k]^2/(Pochhammer[1/4 - m, k] Pochhammer[3/4 - m, k]), {m, ∞, 1}]
// Normal // FullSimplify[#, k ∈ Integers] &
(* (E^((1/96)/m) (-1 + 96 m))/(96 m) *)
Limit[%, m -> ∞]
(* 1 *)
Hence, in this limit f65
reduces to
f43 = 2^(3/2 - 2 m) Sqrt[m/π] Sum[
Pochhammer[1/4, k] Pochhammer[3/4, k] Pochhammer[-m, k]^2/
(Pochhammer[1/2, k]^2 Pochhammer[1, k])/k!
, {k, 0, ∞}]
(* (2^(3/2 - 2 m) Sqrt[m/π] HypergeometricPFQ[{1/4, 3/4, -m, -m}, {1/2, 1/2, 1}, 1] *)
At this point it is, perhaps, worth noting that
f43 /. m -> 100000 // N[#, 10] &
(* 0.6366197724 *)
still is equal to 2/π
to 10
significant figures. Interestingly, the terms in the f43
Sum
can be further simplified by
t43s = FullSimplify[2^(3/2 - 2 m) Sqrt[m/π] Pochhammer[1/4, k] Pochhammer[3/4, k]
Pochhammer[-m, k]^2/(Pochhammer[1/2, k]^2 Pochhammer[1, k])/k!,
k ∈ Integers && m ∈ Integers]
(* (2^(3/2 + 2 k - 2 m) Sqrt[m]
Gamma[1/2 + 2 k] Pochhammer[-m, k]^2)/(π Gamma[1 + 2 k]^2) *)
but summing over t43s
just returns f43
. I know of no exact simplifications of f43
in the large m
limit, although one may exist.
Because Pochhammer[-m, k]
vanishes for k > m
and m
a positive integer, it is possible to evaluate and plot every non-vanishing t43s
. For instance,
ListPlot[Table[t43s /. m -> 1000, {k, 0, 1000}], DataRange -> 1, PlotRange -> All]

A quite good fit to this curve is given by
4/π^(3/2) m^(-1/2) Exp[-4 m ((k - m/2)/m)^2]`
The coefficient 4/π^(3/2) m^(-1/2)
can be obtained from
Assuming[m > 0 && m ∈ Integers, Series[(2^(3/2 + 2 k - 2 m) Sqrt[m] Gamma[1/2 + 2 k]
Pochhammer[-m, k]^2)/(π Gamma[1 + 2 k]^2) /. k -> m/2, {m, ∞, 0}]] // Normal
for even m
. An expansion of t43s
about k = m/2
then gives the width 2/Sqrt[m]
. The resulting agreement is very good, as can be seen from
With[{m = 1000}, ListLogPlot[{Table[(2^(3/2 + 2 k - 2 m) Sqrt[m]
Gamma[1/2 + 2 k] Pochhammer[-m, k]^2)/(π Gamma[1 + 2 k]^2), {k, 0, m}],
Table[(2 m)^-(1/2) Exp[-4 m ((k - m/2)/m)^2], {k, 0, m}]},
DataRange -> 1, PlotRange -> All]]

(t43s
is blue, the fitted function orange.) The fitted function can be integrated to give
Assuming[ m > 0, Integrate[4/π^(3/2) m^(-1/2) Exp[-4 m ((k - m/2)/m)^2], {k, -∞, ∞}]]
(* 2/π *)
as desired.
Addendum: Improved Proof Based on Pochhammer Substitution
The barrier to expanding f43
is Pochhammer[-m, k]
, which is not continuous for positive integer m
and k
. However, even though Mathematica appears not to know it,
Assuming[m >= k >= 0 && (m | k) ∈ Integers,
Pochhammer[-m, k]^2 == Pochhammer[m + 1 - k, k]^2]
is True
, as can be seen empirically by trying numerous values of m
and k
. Furthermore, making this substitution into the expression above for f43
gives the same result as before. So, let us focus on the alternative representation,
t43a = (2^(3/2 - 2 m) Sqrt[m] Pochhammer[1/4, k] Pochhammer[3/4, k]
Pochhammer[m + 1 - k, k]^2)/(Sqrt[π] k! Pochhammer[1/2, k]^2 Pochhammer[1, k])
Because it is apparent from the first plot above that t43a
is large only near k = m/2
, t43a
should be expanded for large m
with k = a m
.
(Assuming[0 < a < 1, Series[t43a /. k -> a m, {m, ∞, 1}]] //
Normal) // FullSimplify[#, m > 0] &
(* (4^-m (1 - a)^(-1 + 2 (-1 + a) m) a^(-1 - 2 a m))/(Sqrt[m] π^(3/2)) *)
Exp[(Series[Log[%], {a, 1/2, 2}] // Normal) /. a -> k/m]
(* (4 E^((-(1/2) + k/m)^2 (4 - 4 m)))/(Sqrt[m] π^(3/2)) *)
Limit[Integrate[%, {k, -∞, ∞}, Assumptions -> m > 1], m -> ∞]
(* 2/π *)
which provides a superior derivation, one not involving curve-fitting.