Vtyjkyuk

After numerical analysis it seems that

$$
fracpi^332=1-sum_k=1^inftyfrac2k(2k+1)zeta(2k+2)4^2k+2
$$

Could someone prove the validity of such identity?

Yes, we can prove it. We can change the order of summation in

$$beginalign
sum_k=1^infty frac2k(2k+1)zeta(2k+2)4^2k+2
&= sum_k=1^infty frac2k(2k+1)4^2k+2sum_n=1^infty frac1n^2k+2\
&= sum_n=1^infty sum_k=1^infty frac2k(2k+1)(4n)^2k+2\
&= sum_n=1^infty r''(4n),
endalign$$

where, for $lvert zrvert > 1$, we define

$$r(z) = sum_k=1^infty frac1z^2k = frac1z^2-1 = frac12left(frac1z-1 - frac1z+1right).$$

Differentiating yields $r''(z) = frac1(z-1)^3 - frac1(z+1)^3$, so

$$1 - sum_k=1^infty frac2k(2k+1)zeta(2k+2)4^2k+2 = sum_nu = 0^infty frac(-1)^nu(2nu+1)^3,$$

and the latter sum is by an earlier answer using the partial fraction decomposition of $dfrac1cos z$:

$$sum_nu=0^infty frac(-1)^nu(2nu+1)^3 = - fracpi^332 E_2 = fracpi^332.$$

Using the formula for a geometric series,
$$
beginalign
sum_k=1^inftyfrac1x^2k
&=frac1x^2-1\
&=frac12left(frac1x-1-frac1x+1right)tag1
endalign
$$
Differentiating $(1)$ twice,
$$
sum_k=1^inftyfrac2k(2k+1)x^2k+2
=frac1(x-1)^3-frac1(x+1)^3tag2
$$
Changing the order of summation and applying $(2)$,
$$
beginalign
1-sum_k=1^inftyfrac2k(2k+1)zeta(2k+2)4^2k+2
&=1-sum_k=1^inftysum_j=1^inftyfrac2k(2k+1)(4j)^2k+2\
&=1-sum_j=1^inftysum_k=1^inftyfrac2k(2k+1)(4j)^2k+2\
&=1-sum_j=1^inftyleft(frac1(4j-1)^3-frac1(4j+1)^3right)\
&=1-sum_n=1^inftyfrac(-1)^n-1(2n+1)^3\
&=sum_n=0^inftyfrac(-1)^n(2n+1)^3tag3
endalign
$$
The sum in $(3)$ can be generalized as
$$
beta(s)=sum_n=0^inftyfrac(-1)^n(2n+1)^stag4
$$
and is known as the Dirichlet beta function. As shown below, $beta(3)=dfracpi^332$.

We can develop a recurrence for $beta(2k+1)$. First, the generating function for $beta(2k+1)$ is
$$
beginalign
f(x) &= sum_k=0^infty beta(2k+1) x^2k+1\
&= sum_n=0^infty sum_k=0^infty (-1)^nleft(fracx2n+1right)^2k+1\
&= sum_n=0^infty (-1)^nfracfracx2n+11-left(fracx2n+1right)^2\
&= fracx2 sum_n=0^infty (-1)^n left(frac12n+1+x+frac12n+1-xright)\
&= fracx2 sum_n=-infty^+infty(-1)^n frac12n+1+x\
&= fracx4 sum_n=-infty^+infty(-1)^n frac1n+tfrac1+x2\
&= fracx4 pi cscleft(pitfrac1+x2right)\[3pt]
&= fracpi4 x secleft(fracpi2xright)tag5
endalign
$$
where we use $(7)$ from this answer to get
$$
beginalign
sum_n=-infty^+inftyfrac(-1)^nn+z
&=sum_n=-infty^+inftyfrac22n+z-sum_n=-infty^+inftyfrac1n+z\[3pt]
&=picot(pi z/2)-picot(pi z)\[9pt]
&=picsc(pi z)tag6
endalign
$$
We can use equation $(5)$ to develop a recurrence relation:
$$
beginalign
fracpi4 x
&= cosleft(fracpi2 xright) f(x)\
&= sum_n=0^inftysum_k=0^n (-1)^k frac(fracpi2 x)^2k(2k)!;beta(2n-2k+1)x^2n-2k+1\
&= sum_n=0^inftysum_k=0^n frac(-pi^2/4)^k(2k)!;beta(2n-2k+1)x^2n+1tag7
endalign
$$
For $n=0$, we can use the arctan series to get
$$
beta(1) = fracpi4tag8
$$
and for $ngt0$, $(7)$ gives
$$
beta(2n+1) = -sum_k=1^n frac(-pi^2/4)^k(2k)!;beta(2n-2k+1)tag9
$$
Recursion $(9)$ yields
$$
beginalign
beta(1)&=fracpi4\
beta(3)&=fracpi^332\
beta(5)&=frac5pi^51536\
beta(7)&=frac61pi^7184320\
beta(9)&=frac277pi^98257536\
beta(11)&=frac50521pi^1114863564800\
beta(13)&=frac540553pi^131569592442880\
beta(15)&=frac199360981pi^155713316492083200\
beta(17)&=frac3878302429pi^171096956766479974400\
beta(19)&=frac2404879675441pi^196713375410857443328000
endalign
$$

Starting from the Laurent series of the cotangent function:
$$pi zcot left( pi z right) =1-2,sum _k=0^infty zeta
left( 2,k+2 right) z^2k+2 tag1$$
apply the differential operator:
$$hatD=z^2dfracd^2dz^2-2zdfracddz+2 tag2$$
to get:
$$z^3pi ^3cot left( pi z right) left( 1+ cot
left( pi z right) ^2 right) =1-sum _k=0^infty 2k left( 2k+1 right)
,zeta left( 2,k+2 right) z^2k+2tag3$$
which, by the ratio test, has a radius of convergence of $|z|<1$. Then from:
$$z=dfrac14, quad cotleft(dfracpi4right)=1 tag4$$
we have:
$$dfracpi ^332=1-sum _k=0^infty frac 2k left( 2,k+1
right) zeta left( 2,k+2 right) 4^2k+2tag5$$

$$
beginalign
S, &=,sum_k=1^inftyfrac2k,(2k+1),zeta(2k+2)4^2k+2 ,=,sum_k=1^inftyfrac(2k+1)!,zeta(2k+2)(2k-1)!,4^2k+2 \[4mm]
&=,sum_k=1^inftyfrac(1/4)^2k+2(2k-1)!,int_0^inftyfracx^2k+1e^x-1,dx ,=,4^-3int_0^inftyfracx^2e^x-1,left(sum_k=1^inftyfrac(x/4)^2k-1(2k-1)!right),dx \[4mm]
&=,4^-3int_0^inftyfracx^2e^x-1,sinhleft(fracx4right),dx ,=,frac4^-32int_0^inftyfracx^21-e^-x,left(Large e^-frac34x,-,Large e^-frac54xright),dx \[4mm]
&=,4^-3left[zetaleft(3,frac34right),-,zetaleft(3,frac54right)right] ,=,1-4^-3left[zetaleft(3,frac14right),-,zetaleft(3,frac34right)right] \[4mm]
&=,1-beta(3) ,=,colorred1-fracpi^332 \
endalign
$$

$,zeta(s,q) ,,$ : Hurwitz Zeta Function

$,beta(s)quad,$ : Dirichlet Beta Function

First, Mathematica confirms this. For the proof, I would assume that expanding out the zeta and then changing the order of summation will do the trick (you will get sums of the sort
$sum_k=1^infty 2k(2k+1)frac1(4 i)^2k+2,$ which are easy to evaluate (by integrating the sum of the geometric series a couple of times):

$$frac2 left(48 i^2+1right)left(16 i^2-1right)^3$$