= SIR final removed fraction
{c}
{title2=$r_\infty=1-s_\infty$}
For the closed <SIR model> with $\gamma>0$, the <infectious compartment>'s <fraction> approaches $i_\infty=0$. One proof outline uses the bounded, increasing <removed compartment>'s <fraction>: since $r(t)-r_0=\gamma\int_0^t i(u)\,du$, the total time-integrated $i$ is finite. The bounded <derivative> $di/dt$ from the <SIR model equations> prevents indefinitely recurring spikes, so $i(t)$ tends to zero.
For $s_0>0$, the <final size relation for an epidemic> follows by dividing the <SIR model equations> for $s$ and $r$ while $i>0$, giving $ds/dr=-(\beta/\gamma)s$. <Integration> then yields
$$
\ln\frac{s(t)}{s_0}=-R_0\bigl(r(t)-r_0\bigr),\qquad
\ln\frac{s_\infty}{s_0}=-R_0\bigl(r_\infty-r_0\bigr),\qquad
r_\infty=1-s_\infty.
$$
This also follows from the <SIR first integral> and <conservation of population>. For the stipulated <initial value problem>, solve
$$
\ln\frac{s_\infty}{0.99}=-3(1-s_\infty).
$$
The <bisection method> gives
$$
s_\infty\approx0.05879736480,\qquad
r_\infty\approx0.94120263520.
$$
About $5.88\%$ of the original population remains in the <susceptible compartment>. The $94.12\%$ is the final <removed compartment>'s <fraction>, including the initially $1\%$ in the <infectious compartment>. The newly acquired <infection> during the modelled period accounts for $s_0-s_\infty\approx0.93120263520$ of the total population, not $1-s_\infty$. Dividing by $s_0$ instead gives the proportion of initially <susceptible compartment> members who acquire <infection>; its denominator must be stated.
At the <SIR infection peak>, the flow into the <infectious compartment> only balances its outflow; it does not stop. Further <infection> explains why the final <susceptible compartment>'s <fraction> is well below $1/3$. Also $s_\infty=s_0e^{-R_0(r_\infty-r_0)}>0$ for $s_0>0$ and finite $R_0$: the closed <SIR model> does not require everybody to acquire <infection> before its <infectious compartment> empties in the long-time <limit>.
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