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Write $V(x)=F(\pi ,x)$. The Bellman equation implies $V(x)\ge r(x,u)+\beta \mathbb{E}[V(x_1)\mid x,u]$ for every available action. Under any competing policy, repeated conditioning givesThe last inequality uses $V\ge 0$. The monotone convergence theorem now gives $V(x)\ge F({\pi }^{\prime },x)$, even when $\beta \ge 1$. If $V(x)=\infty$ the comparison is immediate; otherwise the nonnegative conditional bounds justify each finite induction step. This proves Bellman comparison with nonnegative rewards and superunit discount without a vanishing-terminal-term assumption.

For the unit-bet policy in this gambler's ruin problem, put $\beta =\sqrt{9/8}=3/(2\sqrt{2})$. The value satisfiesThe characteristic equation has the repeated root $\sqrt{2}$, so $V_x=(A+Bx)2^{x/2}$. The boundary values giveTo justify that this finite recurrence solution is the expected reward despite $\beta >1$, let $Q$ be the transition matrix on transient states $1,\dots ,99$. The matrix $\beta Q$ is diagonally similar, with diagonal factors $2^{x/2}$, to the symmetric tridiagonal matrix with off-diagonal entries $1/2$. Its eigenvalues are $\cos (j\pi /100)$, $1\le j\le 99$, with absolute value below one. Thus its Neumann series converges and yields exactly the absorbed-payoff expectation and the unique recurrence solution.

When $p_2=1/4$, at wealth $2$ bet $2$ once and then follow the unit-bet policy. Its expected value is $\beta V_4/4=\beta V_2>V_2$, since $V_4=4V_2$. Therefore the unit-bet policy is not optimal.