= Solution
For the <Poisson process> of claims and <relative safety loading> $\rho$, the premium rate in the <classical risk model> is $c=(1+\rho)\lambda\mu$. Dividing the <adjustment coefficient> equation $\lambda(M(r)-1)=cr$ by $\lambda$ explains why the arrival rate cancels. Interpret $r_\infty$ as the upper endpoint of the finite <moment-generating function> domain, so $M(r)<\infty$ for $0\leq r<r_\infty$ and its stated divergence occurs at that endpoint.
Set $F(r)=M(r)-1-(1+\rho)\mu r$. Then
$$
F(0)=0,\qquad F'(0)=-\rho\mu<0,\qquad
F''(r)=\mathbb E[X_1^2e^{rX_1}]>0\quad(0\leq r<r_\infty).
$$
Differentiation is valid inside the finite-transform domain, and its positive neighbourhood gives finite <moments> of every order. Thus $F$ is <strictly convex> and negative just to the right of zero. If $r_\infty<\infty$, divergence of $M(r)$ makes $F(r)\to\infty$ at the endpoint. If $r_\infty=\infty$, choose $a>0$ with $\mathbb P(X_1\geq a)>0$; then $M(r)\geq\mathbb P(X_1\geq a)e^{ar}$, which eventually exceeds every <linear function>. Again $F(r)\to\infty$. The <intermediate value theorem> gives a positive zero.
There is at most one positive zero: <strict convexity> and $F(0)=0$ make the secant slope $F(r)/r$ strictly increasing for $r>0$. It starts at $-\rho\mu$ and crosses zero exactly once. Hence
$$
\boxed{\text{There is a unique }R\in(0,r_\infty)\text{ with }M(R)-1=(1+\rho)\mu R.}
$$
This proves the <secant-slope existence criterion for an adjustment coefficient> in the present setting.
Put $m_j=\mathbb EX_1^j$. For $u>0$, the positive remainder of the <exponential series> gives $e^u>1+u+u^2/2$. Apply it to $u=RX_1$ and take <expectations>:
$$
(1+\rho)\mu R=M(R)-1>\mu R+\frac{m_2R^2}{2}.
$$
Dividing by $R>0$ yields
$$
\boxed{R<r_1=\frac{2\rho\mu}{m_2}.}
$$
Keeping the cubic term similarly gives
$$
\rho\mu>\frac{m_2R}{2}+\frac{m_3R^2}{6}.
$$
The <polynomial> $g(r)=m_3r^2+3m_2r-6\rho\mu$ is strictly increasing for $r\geq0$, is negative at zero and tends to infinity. Let $r_2$ be its unique positive zero. The last inequality says $g(R)<0$, so
$$
\boxed{R<r_2=\frac{\sqrt{9m_2^2+24\rho\mu m_3}-3m_2}{2m_3},\qquad
g(r_2)=0.}
$$
Finally, $3m_2r_1=6\rho\mu$, whence $g(r_1)=m_3r_1^2>0$. Monotonicity gives $r_2<r_1$. Thus the requested <upper bounds> can in fact be sharpened to the strict ordering \b[$0<R<r_2<r_1$]. The bounds themselves need not lie in the finite-transform domain; the argument only evaluates the actual transform at $R$ and <polynomial> <moments> elsewhere. These are <polynomial moment bounds for the adjustment coefficient>.
For rate-$\alpha$ <exponential distribution> claims,
$$
\mu=\frac1\alpha,\qquad m_2=\frac2{\alpha^2},\qquad m_3=\frac6{\alpha^3},\qquad
M(r)=\frac\alpha{\alpha-r}\quad(r<\alpha).
$$
The positive root satisfies $1/(\alpha-R)=(1+\rho)/\alpha$ after dividing out $R$, so
$$
\boxed{R=\frac{\alpha\rho}{1+\rho},\qquad
r_1=\alpha\rho,\qquad
r_2=\frac\alpha2\left(\sqrt{1+4\rho}-1\right).}
$$
The cubic-bound equation reduces to $r_2^2+\alpha r_2-\rho\alpha^2=0$, selecting the displayed positive root. In particular $R<\alpha$, as required by the transform domain, for every $\rho>0$.
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