Correct answer: (C) V is exponential of mean 5; (D) e^(−V/5) is uniform on (0,1)
Explanation
For z≥0, P(Z_n>z)=(1−1/(5n))^(floor(nz)+1)→e^(−z/5). Thus V is exponential of mean 5. For 0<u<1, P(e^(−V/5)≤u)=P(V≥−5ln u)=u, proving the uniform transform. **C,D**.
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