Evaluating a sum from the Berkeley Math tournament

[@slavinojunepri7648]

Fantastic

[@dwightswanson3015]

Multiplying by x, taking the derivative twice of a sum -- new (to me) methods

[@Andrew-II]

5*e from the last result minus 5*e from the original sum equal to 0? sum [n from 0 to inf] ((n^3-5)/n!) = 0 Heh, it looks weird 🧐

[@thefeynmantechnique]

I'm not sure what you mean.

[@Andrew-II]

@@thefeynmantechnique I subtracted from the resulting sum (n^3/n!), which is equal to 5*e, the fivefold value of the original sum 1/n!, which is equal to e. It should turn out to be 0, right?

[@Andrew-II]

However, I've built a discrete graph, and it looks like it's true.

[@thefeynmantechnique]

@@Andrew-II yes, sum from 0 to infinity of (n^3-5)/n! is 0.