Question: As before, note that X k=n k 1 n 1 pnqkn = pn X l=0 n +l 1 l

As before, note that

∞X k=n



k − 1 n − 1



pnqk−n = pn ∞X l=0



n +l − 1 l



ql where l = k − n

= pn ∞X l=0



−n l



(−q)l

= pn(1 − q)−n = 1, using the binomial expansion of (1 − q)−n, see Theorem A.3.

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