Sums of reciprocals of powers of the imaginary part of the nontrivial zeros of the Riemann zeta function

Question

To investigate the relations between Appell / Jensen polynomials and the real-valued function for real $t$

$$\Omega(t) = \xi(1/2+it)/\xi(1/2),$$

where $\xi(s)$ is the Landau Riemann xi function, I need reasonable approximations, say to 3 significant digits, of

$$ Tr_{2n}= \sum_{k=0}^\infty 1/(z_k)^{2n}$$

for $n=1,2,3,4$ where the $z_k$ are the imaginary parts of the nontrivial zeros of the Riemann zeta function above the real axis. (Assume the RH is true, of course.)

I don't have access to Mathematica nor Maple, so help would be appreciated.

Edit (June 5, 2020):

To allay any further doubts about the convergence of $Tr_2$:

Titchmarsh in his classic book On the Theory of the Riemann Zeta Function has, on p. 18, Eqn 2.1.14

$$\Xi(z)= \xi(1/2+iz),$$

and on p. 30 he states that it is an even integral function of order 1, whose exponent of convergence is 1. "Hence $\Xi(z)$ has an infinity of zeros, whose exponent of convergence is 1. The same is true of $\xi(s).$" In his Theory of Functions on p. 249 is

Theorem 8.22: If $r_1, r_2...$ are the moduli of the roots of $f(z)$, then the series $\sum 1/r^{\alpha}$ is convergent if $\alpha > \rho.$

$\rho$ in an earlier paragraph is called the order of the integral function $f(z)$.

The absolute contribution of a zero of $\Omega$, $a+ib$, and its complex conjugate to the sum of the inverse squares of the zeros is $2(a^2-b^2)/(a^2+b^2)^2=2\cos(2\theta)/r^2$ with $\theta=0$ for the real zeros. This is less than $2/r^{\alpha}$ for $0< \alpha < 2$, so the trace of the paired inverse squares, even including any complex zeros if they were to be found, is absolutely convergent

Answer 1

Just to make things visible and to help towards a hypothese (well aware of the comment of Gerry Myerson): I've somewhere found online the set of first 100 000 values of the imaginary parts to 8 digits. Partial sums in increasing segments of $2^k$ are the following:

   partial sums of sum(i=1,2*2^k-1, 1/(imag(root_i))^(2n)   

   k     n=1             n=2                 n=3                  n=4
  -----------------------------------------------------------------------------------------
   0  0.00500524412341  0.0000250524687349  0.000000125393721912  0.000000000627626189713
   1  0.00886667676944  0.0000327283874871  0.000000141065528135  0.000000000660375162896
   2   0.0121739737744  0.0000356030857633  0.000000143677509519  0.000000000662837761448
   3   0.0150045293315  0.0000366728779872  0.000000144108076453  0.000000000663021051125
   4   0.0172998726364  0.0000370263290034  0.000000144166228966  0.000000000663031185306
   5   0.0190794372367  0.0000371333200692  0.000000144173144661  0.000000000663031660997
   6   0.0203971381635  0.0000371628546828  0.000000144173860457  0.000000000663031679531
   7   0.0213326190435  0.0000371703457908  0.000000144173925660  0.000000000663031680139
   8   0.0219725965445  0.0000371721091123  0.000000144173930966  0.000000000663031680156
   9   0.0223966100709  0.0000371724981445  0.000000144173931358  0.000000000663031680157
  10   0.0226699376045  0.0000371725793391  0.000000144173931384  0.000000000663031680157
  11   0.0228420530479  0.0000371725955004  0.000000144173931386  0.000000000663031680157
  12   0.0229482871375  0.0000371725985893  0.000000144173931386  0.000000000663031680157
  13   0.0230127402289  0.0000371725991596  0.000000144173931386  0.000000000663031680157
  14   0.0230512683394  0.0000371725992617  0.000000144173931386  0.000000000663031680157
  15   0.0230740041036  0.0000371725992795  0.000000144173931386  0.000000000663031680157

update corrected wrong upper index in sum-expression
update2 corrected the mention in the partial sums protocol, that the sums are of course from the powers of the imaginary part of root_i alone.