We know that $\sum_{k=0}^n a^k \frac{n^{\underline k}}{k!} = (1+a)^n$. Is there a known (preferably closed) form for $\sum_{k=0}^n a^k/k!$ ?
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Letting $n\to \infty$ yields: $$ \sum_{k=0}^\infty \dfrac{a^k}{k!} = e^a $$ – Adriano Jul 04 '13 at 03:30
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It can be written in terms of the incomplete Gamma function: $\dfrac{e^a}{n!} \Gamma(n+1,a)$.
Robert Israel
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