Let $\{u_n\}$ be a sequence in which the first $k$ terms are given and each term after $k$th term is the geometric mean of the preceding $k$ terms. $$u_{k+1}=(u_1u_2u_3...u_k)^{\frac 1k}$$ $$u_{k+2}=(u_2u_3u_4...u_{k+1})^{\frac 1k}$$ $$...$$ $$u_n=(u_{n-k}u_{n-(k-1)}...u_{n-1})^{\frac 1k}$$ As $n\to\infty$, what does $\{u_n\}$ converge to?
This is an attempt to generalize this question Finding the limit of the sequence with $a_1=a$, $a_2=b$, and $a_{n}=\sqrt{a_{n-1} a_{n-2}}$
Using the ideas that every term after the first $k$th term is the product of the powers of the first $k$ terms and that the index of each power is the arithmetic mean of the indices of that same-base power in the preceding $k$ terms, we can turn this problem into finding Limit of sequence in which each term is the average of its preceding k terms.
With the answer to the arithmetic mean sequence problem, we can just substitute the initial indices into the formula and find that the sequence converges to $u_1^{\frac {1}{S}}u_2^{\frac {2}{S}}u_3^{\frac {3}{S}}...u_k^{\frac {k}{S}}$, where $S=\sum_{i=1}^{k} i = \frac {k(k+1)}{2}$.