[THOUGHT] We need to solve the problem step by step, using Python to compute necessary values. The problem has multiple parts I–V, but ultimately we need to compute 𝒦 = 999(p² - 1)(-3r)(-D₂)(S₂/T). We are given D₂ = -2/3. We need p (order of pole at z=0), r (residue at z=0), S₂ (sum from IV), T (sum from V with N=1). We'll compute these using Python.

First, understand f(z) = cot(z)/z². cot(z) = cos(z)/sin(z). Poles occur when denominator is zero: sin(z)=0 or z²=0. sin(z)=0 at z = nπ for integer n. Also z=0 is a pole because cot(z) has a simple pole at 0? Actually cot(z) = cos(z)/sin(z). At z=0, sin(0)=0, cos(0)=1, so cot(z) has a simple pole at 0. Then f(z) = cot(z)/z² has a pole at z=0 of order? Let's examine: cot(z) ~ 1/z - z/3 - z³/45 - ... (Laurent expansion). So cot(z)/z² ~ 1/z³ - 1/(3z) - ... So pole at 0 is order 3? But careful: the series expansion: cot z = 1/z - z/3 - z³/45 - ... So f(z) = (1/z - z/3 - z³/45 - ...)/z² = 1/z³ - 1/(3z) - ... So indeed pole at z=0 is order 3 (since highest negative power is z⁻³). So p = 3.

Now residue at z=0: coefficient of 1/z in Laurent expansion. From above: -1/3. So r = -1/3.

But we should verify using Python with sympy to compute residue.

Now S₂: from IV, using residue theorem on square contour, we should get sum_{n=1}^∞ 1/n² = π²/6. That's a known result. But we need to compute via the method described. However, we can trust known value: S₂ = π²/6.

Now T: from V, general formula for sum_{n=1}^∞ 1/n^{2N} in terms of D_m. For N=1, T should equal S₂, so T = π²/6 as well. But we need to derive expression.

Let's derive general formula. g(z) = cot(z)/z^{2N}. Poles: at z = nπ, n integer, n≠0, simple poles? Actually cot(z) has simple poles at nπ, and denominator z^{2N} is analytic nonzero there, so simple poles at z = nπ, n≠0. At z=0, cot(z) has simple pole, so g(z) has pole of order 2N+1? Let's check: cot(z) ~ 1/z - z/3 - ... So g(z) ~ (1/z - z/3 - ...)/z^{2N} = 1/z^{2N+1} - 1/(3 z^{2N-1}) - ... So pole order at 0 is 2N+1.

But we only need T for N=1, which is sum_{n=1}^∞ 1/n². The general formula likely involves D_{2N} or something. D_m = lim_{z→0} d^m/dz^m (z cot z). Note that z cot z is even and analytic at 0, with expansion: z cot z = 1 - (1/3) z² - (1/45) z⁴ - ... So D_m are coefficients times factorial? Actually D_m = m! times coefficient of z^m in expansion of z cot z? Because Taylor expansion: z cot z = Σ_{k=0}∞ a_k z^k, then D_m = m! a_m. Indeed D_2 = -2/3, so a_2 = -1/3, D_2 = 2! * (-1/3) = -2/3 matches.

Now residue at z=0 for g(z): we need coefficient of 1/z in Laurent expansion of g(z). That's related to D_{2N-1}? Let's compute for N=1: we already have residue -1/3. For general N, residue at 0 is -D_{2N-2}/(2N-1)!? Not sure.

But we can compute using Python symbolic expansion.

We'll compute p, r, S₂, T using Python.

First, compute p: order of pole at 0 for f(z). Use sympy series.

Second, compute residue at 0 for f(z).

Third, compute S₂: we can compute sum_{n=1}^∞ 1/n² = π²/6 numerically.

Fourth, compute T: from general formula in V, we need to derive expression for sum_{n=1}^∞ 1/n^{2N} in terms of D_m. Then set N=1.

We can attempt to derive using residue theorem on same square contour. The integral around square tends to 0 as M→∞ for f(z) and g(z) because cot(z) is bounded on the contour away from poles? Actually cot(z) is bounded on vertical and horizontal lines not passing through poles? For large |z|, cot(z) → ±i? Actually cot(z) = i (e^{iz}+e^{-iz})/(e^{iz}-e^{-iz}). For z = x+iy, large y, cot(z) → i sign(y). So bounded. So integral on each segment tends to 0 as R→∞? Possibly.

But we can compute using known results: sum_{n=1}∞ 1/n^{2N} = (-1)^{N+1} B_{2N} (2π)^{2N} / (2 (2N)!) where B are Bernoulli numbers. And D_m are related to Bernoulli numbers via expansion of z cot z = Σ_{k=0}∞ (-1)^k B_{2k} (2z)^{2k}/(2k)!.

Thus D_{2N} = (2N)! * coefficient of z^{2N} in z cot z = (2N)! * (-1)^N B_{2N} 2^{2N} / (2N)! = (-1)^N B_{2N} 2^{2N}. So B_{2N} = (-1)^N D_{2N} / 2^{2N}.

Then sum_{n=1}∞ 1/n^{2N} = (-1)^{N+1} B_{2N} (2π)^{2N} / (2 (2N)!) = (-1)^{N+1} [(-1)^N D_{2N} / 2^{2N}] * (2π)^{2N} / (2 (2N)!) = (-1)^{2N+1} D_{2N} π^{2N} / ( (2N)! )? Let's compute:

(-1)^{N+1} * (-1)^N = (-1)^{2N+1} = -1. So sum = - D_{2N} / 2^{2N} * (2π)^{2N} / (2 (2N)!) = - D_{2N} * (2π)^{2N} / (2^{2N+1} (2N)!) = - D_{2N} * π^{2N} * 2^{2N} / (2^{2N+1} (2N)!) = - D_{2N} π^{2N} / (2 (2N)!).

Thus sum_{n=1}∞ 1/n^{2N} = - D_{2N} π^{2N} / (2 (2N)!).

For N=1, D_2 = -2/3, so sum = - (-2/3) π² / (2 * 2!) = (2/3) π² / 4 = (2π²)/(12) = π²/6. Correct.

Thus T = π²/6.

So S₂/T = 1.

Now compute p=3, r=-1/3, D_2=-2/3.

Now compute 𝒦 = 999 (p² - 1) (-3r) (-D₂) (S₂/T).

Plug numbers: p² - 1 = 9 - 1 = 8. -3r = -3 * (-1/3) = 1. -D₂ = -(-2/3) = 2/3. S₂/T = 1.

So product = 999 * 8 * 1 * (2/3) * 1 = 999 * 8 * 2/3 = 999 * 16 / 3 = 333 * 16 = 5328.

Let's verify with Python.

[/THOUGHT] [PYTHON] import sympy as sp

Define symbols

z = sp.symbols('z')

Define f(z)

f = sp.cot(z) / z**2

Find poles and orders

Poles where denominator zero: sin(z)=0 or z=0

Use series to find order at 0

series_f = sp.series(f, z, 0, 5) print("Series expansion of f(z) around 0:") print(series_f)

Determine order: highest negative power

Let's extract leading term

order = -sp.Poly(series_f.removeO().as_leading_term(z, x0=0), z).degree() print("Order of pole at 0:", order)

Compute residue at 0 using sympy

residue_0 = sp.residue(f, z, 0) print("Residue at 0:", residue_0)

Compute S2 = sum_{n=1}^∞ 1/n^2

S2 = sp.nsum(lambda n: 1/n2, (1, sp.oo)) print("S2 =", S2) print("S2 as rational times pi^2:", S2 / sp.pi2)

Compute D2 given

D2 = -sp.Rational(2,3) print("D2 =", D2)

Compute T from general formula for N=1: T = - D2 * pi^2 / (2 * 2!)

T = - D2 * sp.pi**2 / (2 * sp.factorial(2)) print("T =", T) print("T simplified:", sp.simplify(T))

Compute ratio S2/T

ratio = S2 / T print("S2/T =", ratio)

Compute p, r from above

p = order r = residue_0 print("p =", p) print("r =", r)

Compute K

K = 999 * (p**2 - 1) * (-3 * r) * (-D2) * ratio print("K =", K) print("K as integer:", sp.simplify(K)) [/PYTHON]

[THOUGHT] The Python code had an error in extracting the order, but I already know from series expansion that the pole order at 0 is 3 (since cot(z)/z² ~ 1/z³). The residue at 0 is -1/3. S₂ = π²/6, T = π²/6, so S₂/T = 1. Given p=3, r=-1/3, D₂=-2/3, compute 𝒦 = 999*(3²-1)(-3(-1/3))(-(-2/3))1 = 99981*(2/3)1 = 99916/3 = 333*16 = 5328. This matches my earlier calculation. The Python error does not affect the known values. So final answer is 5328.

[/THOUGHT] \boxed{5328}