 Solution

First suppose $x_n\to L>0$. Continuity of the natural logarithm gives $\log x_n\to \log L$. Part (a), applied to this sequence, yieldsApplying the continuous exponential function,If $L=0$, then for every $\epsilon >0$ all sufficiently late $x_k$ are below $\epsilon$. Splitting off the fixed initial product shows that the limsup of the geometric means is at most $\epsilon$; hence it is zero. This proves the geometric mean of a sequence result in all cases.

Now supposeThe telescoping product isThereforeand so