= Solution
If $T>n$, either $T_R\leq n$, or both stopping times exceed $n$. Therefore the <maximal bound for a nonnegative martingale> and the <Markov inequality> give, for $R\geq1$,
$$
\mathbb P(T>n)
\leq\mathbb P(T_R<\infty)+\mathbb P(T\wedge T_R>n)
\leq\frac1R+\frac{\tau^2R}{\sigma^2n}.
$$
The right-hand side is minimized at $R=\sigma\sqrt n/\tau$. Whenever the claimed bound is nontrivial, $2\tau/(\sigma\sqrt n)<1$, this choice has $R>2$ and is allowed under the corrected overshoot hypothesis. Substituting gives the <square-root tail bound for martingale absorption>:
$$
\boxed{\mathbb P(T>n)\leq\frac{2\tau}{\sigma\sqrt n}\qquad(n\geq1).}
$$
When the displayed upper bound is at least one, the inequality follows from $\mathbb P(T>n)\leq1$. \b[This handles every $n$ without using the impossible small-threshold hypothesis.]
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