= Solution
For $t\geq0$, the geometric survival <probability> gives
$$
\Pr(T_k>t)=\Pr(S_1>kt)=\left(1-\frac1k\right)^{\lfloor kt\rfloor}\longrightarrow e^{-t}.
$$
For $t<0$, the distribution functions are zero, as is the limiting <exponential distribution> function. Since that limiting distribution function is continuous everywhere, these limits prove
$$
\boxed{T_k\xrightarrow{d}\operatorname{Exp}(1).}
$$
The parameter is rate one, so the limiting mean is one, consistent with $\mathbb E(S_1/k)=1$.
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