This would be equivalent to sum of x^3 over x! too right? (Since you're just multiplying numerator and denominator by x)
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Right, what I'm saying is if you multiply the top and bottom by (n+1), it becomes (n+1)^3/(n+1)! [or more simply, n^3 / n!]
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Yeah you’re right, and quite frankly sum(n^3/n!) looks nicer and really blew my mind.
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