Oh, hey, sorry about the delay. For some reason I didn't get notified of your response. If I hadn't happened to manually check out of curiosity, it may have gone forever unnoticed. I'll have to double check my notification settings. Anyway... what you've done so far is good. You've successfully demonstrated that part of the problem involves a geometric series. More specifically, what you have there is the formula for the expected value of a geometric distribution, which models the probability of getting a "success" on the
kth trial. Just for convenience sake, let's define a few values. Let
p=4π and let
q=1−p=43π. We then have:
S=k=1∑∞k⋅qk−1⋅p=pq+2pq2+3pq3+…
If we multiply this through by
q we get:
q⋅S=k=1∑∞k⋅qk⋅p=pq2+2pq3+3pq4+…
And hence:
S−q⋅S=pq+(2pq2−pq2)+(3pq3−2pq3)+(4pq4−3pq4)+…
Where does this lead you?