= Solution
Write $\tau=t-t_0$, let $R$ be the <scale factor>, $H=\dot R/R$ the <Hubble parameter>, and $\mu=\sqrt{\Lambda/3}$ the asymptotic expansion rate for the positive <cosmological constant>. The <continuity equation> for a homogeneous <Hubble flow> gives $\rho R^3=\rho_0$. Isotropy and the <Poisson equation for Newtonian gravity> give a relative acceleration proportional to displacement from any chosen origin:
$$
\mathbf g=-\frac{4\pi G\rho-\Lambda}{3}\mathbf r.
$$
This uses the local relative <gravitational acceleration>, rather than trying to sum an absolutely convergent potential over an infinite universe. A fluid element with constant comoving coordinate $\mathbf x$ has $\mathbf r=R\mathbf x$. The spatially uniform <pressure> has no gradient, so the <Euler equations for an inviscid fluid> give
$$
\boxed{\ddot R=-\frac{4\pi G\rho_0}{3R^2}+\frac{\Lambda}{3}R.}
$$
Multiply by $2\dot R$ and integrate. With $A=8\pi G\rho_0/3$, the resulting <first integral> is
$$
\boxed{\dot R^2=A/R+\mu^2R^2-\kappa.}
$$
For an expanding branch that reaches arbitrarily large $R$, $\dot R/R\to\mu$; more precisely, $H=\mu+O(R^{-2})$ and the correction to $\ln R-\mu t$ is integrable. Therefore $R\sim C e^{\mu t}$. Positive $\Lambda$ alone does not make every initially expanding branch escape: sufficiently large positive $\kappa$ can give a turning point or a limiting static radius. Under the stated vacuum-dominance inequalities at $R=1$, however, $A/R$ and $|\kappa|$ remain small compared with $\mu^2R^2$ for the future expanding branch, and $R\simeq e^{\mu\tau}$ is the required approximation.
To derive the <linear cosmological density perturbation> equation, write $\delta=\rho'/\rho$ and $\mathbf v=H\mathbf r+\mathbf u$, where $\mathbf u$ is the <peculiar velocity>. At fixed $\mathbf x=\mathbf r/R$, the linearized <continuity equation>, <Euler equations for an inviscid fluid> and perturbation <Poisson equation> are
$$
\dot\delta+R^{-1}\nabla_x\cdot\mathbf u=0,\qquad
\dot{\mathbf u}+H\mathbf u=-R^{-1}\nabla_x(c_s^2\delta+\phi),\qquad
\nabla_x^2\phi=4\pi G R^2\rho\delta.
$$
Taking a <divergence> of the second equation and differentiating the first eliminates the <peculiar velocity>. For a <Fourier mode> of comoving <wavenumber> $k$, this gives
$$
\ddot\varepsilon+2H\dot\varepsilon+\left(\frac{c_s^2k^2}{R^2}-4\pi G\rho\right)\varepsilon=0.
$$
The <adiabatic sound speed> for the four-thirds <polytropic equation of state> is $c_s^2=dp/d\rho\propto\rho^{1/3}=\rho_0^{1/3}/R$, hence $c_s^2=c_0^2/R$. Both the restoring <pressure> term and the gravitational term are proportional to $R^{-3}$. Thus, in the exponential approximation,
$$
\boxed{\ddot\varepsilon+2\mu\dot\varepsilon+\eta^2e^{-3\mu\tau}\varepsilon=0,\qquad
\eta^2=c_0^2k^2-4\pi G\rho_0.}
$$
The sign of $\eta^2$ is the <Jeans instability> distinction, but exponential expansion makes the total subsequent amplification finite.
Remove <Hubble friction> by setting $\varepsilon=e^{-\mu\tau}y$. The standard-form equation is
$$
y''+Q(\tau)y=0,\qquad Q=\eta^2e^{-3\mu\tau}-\mu^2.
$$
Where $Q>0$ and $|Q'|\ll Q^{3/2}$, the <WKB approximation> gives
$$
\varepsilon\simeq e^{-\mu\tau}Q^{-1/4}
\left[A\cos\!\left(\int^\tau\sqrt Q\,ds\right)+B\sin\!\left(\int^\tau\sqrt Q\,ds\right)\right].
$$
Where $Q<0$, put $q=\sqrt{-Q}$; away from a turning point the corresponding <WKB approximation> is
$$
\varepsilon\simeq e^{-\mu\tau}q^{-1/2}
\left[A e^{\int^\tau q\,ds}+B e^{-\int^\tau q\,ds}\right],\qquad |q'|\ll q^2.
$$
For $\eta^2=-s^2<0$, $q=(\mu^2+s^2e^{-3\mu\tau})^{1/2}$ never vanishes. The exponential <integral> can be evaluated explicitly:
$$
\int_0^\tau q(t)\,dt=\frac{2}{3\mu}\bigl[\Psi(q(0))-\Psi(q(\tau))\bigr],\qquad
\Psi(q)=q+\frac{\mu}{2}\ln\frac{q-\mu}{q+\mu}.
$$
Differentiate this expression, using $q'=-3\mu(q^2-\mu^2)/(2q)$, to verify it. Early on, if coefficients change little over the interval considered, the two local exponents for $\varepsilon$ are $-\mu\pm\sqrt{\mu^2+s^2}$: one increases and one decreases. At late times $q\to\mu$ and the displayed <integral> is $\mu\tau+O(1)$. The nominally growing <WKB approximation> branch of $y$ therefore becomes a constant branch of $\varepsilon$, while the other branch decays like $e^{-2\mu\tau}$. The <WKB approximation> is not automatically accurate at all intermediate times; its stated derivative criterion must be checked.
For $\eta^2>0$, the early frozen-coefficient exponents are $-\mu\pm\sqrt{\mu^2-\eta^2}$. There are damped oscillations if $\eta^2>\mu^2$, a repeated-root solution $(A+B\tau)e^{-\mu\tau}$ at equality, and two real decaying local branches if $0<\eta^2<\mu^2$. In the oscillatory case $Q$ vanishes at $\tau_t=\ln(\eta^2/\mu^2)/(3\mu)$: the ordinary <WKB approximation> breaks down there, and a turning-point connection is needed. After that transition the solutions again approach a constant or decay. A positive pressure-minus-gravity restoring term does not imply indefinitely continuing oscillation; its physical <frequency> is redshifted away.
The limiting conclusion does not depend on the accuracy of <WKB approximation>. With $x=2|\eta|e^{-3\mu\tau/2}/(3\mu)$, the equation for $y$ is
$$
x^2y_{xx}+xy_x+\left(\operatorname{sgn}(\eta^2)x^2-\frac49\right)y=0.
$$
For $\eta^2>0$ its basis is $J_{2/3}(x),Y_{2/3}(x)$, the <Bessel functions of the first kind> and <Bessel functions of the second kind>. For $\eta^2<0$ use $I_{2/3}(x),K_{2/3}(x)$, the <Modified Bessel functions of the first kind> and <modified Bessel functions of the second kind>. Their small-$x$ branches are proportional to $x^{2/3}$ and $x^{-2/3}$. Multiplication by $e^{-\mu\tau}$ gives the decaying and constant branches respectively. At $\eta^2=0$ the exact solution is $A+B e^{-2\mu\tau}$. Thus <Bessel density modes during exponential expansion> prove
$$
\boxed{\varepsilon(\tau)\longrightarrow\varepsilon_\infty\quad\text{for every finite initial pair and every real }\eta^2,\ \Lambda>0.}
$$
The same conclusion holds in the full four-thirds perturbation equation on an exponentially escaping background: $(R^2\dot\varepsilon)'=-\eta^2R^{-1}\varepsilon$. Its integrated equation has an absolutely integrable Volterra kernel, bounded by a constant times $e^{-3\mu\tau}$. The <Gronwall inequality> first bounds $\varepsilon$; then $R^{-1}\varepsilon$ is integrable and $\dot\varepsilon=O(R^{-2})$ is integrable. A finite limiting amplitude follows without setting $H$ exactly constant. This argument requires positive $\Lambda$ and the escaping branch; it is not a claim about $\Lambda=0$ or recollapse.
The requested strict increase of that limit needs an initial-velocity qualification. For $\eta^2=-s^2<0$,
$$
\left(e^{2\mu\tau}\dot\varepsilon\right)'=s^2e^{-\mu\tau}\varepsilon.
$$
If $\varepsilon(0)>0$ and $\dot\varepsilon(0)\geq0$, this identity prevents a first loss of positivity and makes the amplitude strictly increase for $\tau>0$. Integrating it twice gives
$$
\varepsilon_\infty=\varepsilon(0)+\frac{\dot\varepsilon(0)}{2\mu}
+\frac{s^2}{2\mu}\int_0^\infty e^{-3\mu\tau}\varepsilon(\tau)\,d\tau
>\varepsilon(0).
$$
This is <monotone Jeans growth with positive cosmological constant>. Without the extra initial condition the printed assertion is false: $\varepsilon=e^{-\mu\tau}I_{2/3}(2s e^{-3\mu\tau/2}/(3\mu))$ is positive at $\tau=0$ and tends to zero. An arbitrarily small multiple remains an infinitesimal <linear cosmological density perturbation>, so linearity does not remove the counterexample.
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