EU. Stacking Cannonballs

Pascal’s triangle, with its host of nifty tricks, provides the surprising solution to last weeks’ puzzle on sequences of averages. 

As a bonus puzzle, not mentioned in the podcast, consider the following variation with a completely different solution: Our sequence starts 

1, 1, …

Now each additional term is twice the average of all the earlier terms, not including the terms immediate predecessor! So, the third term is twice the average of 1, i.e. 2. We have now 1,1,2 …

The fourth term is twice the average of 1 & 1, i.e. 2 and we have 1,1,2,2

Continuing in this way we get 1, 1, 2, 2, 8/3, 3, etc.  The sequence wobbles around, but will grow steadily. But the remarkable thing is that the nth term, divided by n, tends to exactly (1 – 1/e^2)/2, a fact well worth trying to prove!


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