Change of variable in an infinite sum I'm currently trying to understand a derivation from WolframMathWorlds. I got to step 6 where a change of variable happens. You can see the equation here. I understand everything except how they get rid of the infinite sum (because mine is not starting at 0, see below). It could be helpful to take a look at the website but I will try to explain the problem here:
Assume I have a 6 sided dice. I got some generating function defined as:
$$
G_X(s) = E[s^X] = \sum_{x=0}^{\infty} s^x P(X=x)
$$
I got to the point where I have to switch variables:
$$
 k = 6 \cdot i + n + j \\
\frac {1}{6^n} \sum_{i=0}^n {{n} \choose {i}} (-1)^{i} \sum_{j=0}^\infty s^{6\cdot i + n + j}  {n + j - 1 \choose j} = 
\frac {1}{6^n} \sum_{i=0}^n   {{n} \choose {i}}  (-1)^{i} \sum_{k=6 \cdot i + n}^\infty  s^{k} { k - 6 \cdot i - 1 \choose  n - 1 }
$$
Now I can compare the terms and see the similarity to my definition of a generating function. I can use that insight to compute the probabilities as:
$$
P(X=k) = \frac {1}{6^n} \sum_{i=0}^n   {{n} \choose {i}}  (-1)^{i} { k - 6 \cdot i - 1 \choose  n - 1 }
$$
Which makes sense EXCEPT for the part where my sum starts at 6i + n and not at 0. Somehow they still get this result and I do not understand why that is the case, any help? 
P.S. On their website they use a slightly different notation (k=p and i=k for example) but I hope that my questions is quite clear from the code I wrote.

Writing it down made me realize my mistake. It does not matter if the sum starts at 6i +n or at 0 because the first few terms will then just be zero anyway. Maybe somebody finds this useful. 
