Suppose we roll m dice, remove all the dice that come up 1, and roll the rest again. If we repeat this process, eventually all the dice will be eliminated. How many rolls, on average, will we make?
The solution to this problem is $ \sum_{n=1}^{\infty}n \cdot \left( \left(1-\left(\frac{5}{6}\right)^n \right)^m -\left(1-\left(\frac{5}{6}\right)^{n-1}\right)^m\right)$
which can be rewritten to
$\sum_{n=1}^{\infty}\sum_{k=0}^{m}n(-1)^{k} \left (1-\left (\frac{5}{6} \right )^{-k} \right )\binom{m}{k}\left (\frac{5}{6} \right )^{nk}$
For small values of m I can calculate it using Wolfram Alpha, e.g. for m=8
Sum[n*((1 - (5/6)^n)^8 - (1 - (5/6)^(n - 1))^8), {n, 1, Infinity}]
but this does not work for large values of m. Which approximations could I use to calculate a value for m=1000000?