= Solution
All $a_n$ are positive, and $a_1=e^{-1}$. The <logarithm> turns consecutive ratios into differences, producing the finite <telescoping sum>
$$
\sum_{k=1}^{n}u_k=\sum_{k=1}^{n}\bigl(\ln a_{k+1}-\ln a_k\bigr)=\ln a_{n+1}+1.
$$
Thus, already for every finite $n$,
$$
a_{n+1}=\exp\left(-1+\sum_{k=1}^{n}u_k\right).
$$
If the <partial sums> converge to $S$, the <continuity> of the <exponential function> gives
$$
\boxed{\lim_{n\to\infty}a_n=\exp\left(-1+\lim_{n\to\infty}\sum_{k=1}^{n}u_k\right)=e^{-1+S}.}
$$
The shift from $a_{n+1}$ to $a_n$ preserves the <limit of a sequence>. Convergence of the <partial sums> is established in part (f), so no convergence assumption is left unsupported.
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