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by WindyLakeReturn
1162 days ago
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>allowed to calculate past the divergent part My question was more in doubting if the divergent part actually existed. n!x^n will eventually diverge for 0<x<1, regardless of how small x is, but was that really the issue? I thought the problem was more a question if g(n)x^n would diverge when g(n) involves computing n! Feynman diagrams. It can't be automatically assumed that computing n! Feynman diagrams leads to an answer that grows comparable to n!. Or maybe it can, but I didn't see that argument being made anywhere, though I may have missed it. |
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