Start with
ex = 1 + x + x²/2! + x³/3! + ... = ∑n=0∞ xn/n!, which converges everywhere.
Therefore,
x e5x = x ∑n=0∞ (5x)n/n! = ∑n=0∞ 5n xn+1/n!, which also converges everywhere.
David J.
asked 11/03/13
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