I'd like to find the closed form or a quickly converging rewriting of the following n-sum:
$$\left(\frac{1}{6}\right)^n\sum_{x_1=1}^\infty\sum_{x_2=1}^\infty\ldots\sum_{x_n=1}^\infty\left(\frac{5}{6}\right)^{x_1+x_2+\ldots+x_n-n}*max(x_1,x_2,\ldots,x_n)$$
The problem this arises from is when you have n 6-sided dice, roll all of them, set aside the 6's, roll the rest, set aside 6's again and so on until all dice faces are 6's. These sums are the expectation value for the number of rolling rounds to get n 6's.
By numerical summation I got $\approx8.727$ for $n=2$ and $\approx10.555$ for $n=3$ but it's getting impossible to calculate for larger n's and it converges very slowly because of the long tail.
Thanks,
Marton
Standard generating functions technology yields that the $n$th sum $s_n$ is such that, when $n\to\infty$, $$\lim_{n\to\infty}\frac{s_n}{\log n}=\frac1{\log(6/5)}\approx5.485.$$ Is this the kind of result you are after?
For small values of $n$, one could use the exact value as a finite sum in the RHS of the identities $$s_n=\sum_{i\geqslant0}\left(1-(1-(5/6)^i)^n\right)=\sum_{k=1}^n{n\choose k}(-1)^{k+1}\frac{1}{1-(5/6)^k}.$$ For example, $s_1=6$, $s_2=\frac{96}{11}\approx8.73$, $s_3=\frac{10566}{1001}\approx10.56$, as you mentioned, and $s_4=\frac{728256}{61061}\approx11.93$, $s_5=\frac{3698650986}{283994711}\approx13.02$, $s_6\approx13.94$, $s_7\approx14.72$, and so on...
For middle-sized values of $n$, a more eficient formula would be $$s_n=1+\sum_{k=1}^n{n\choose k}(-1)^{k+1}\frac{5^k}{6^k-5^k}.$$