Kummer's transformation for 1F1
Here is the proof of this transformation posted by @infseriesbot ∞∑n=0zn(a)n+1=ez∞∑n=0(−z)n(a+n)
Proof:
Recall the definition of the Pochhammer polynomial (rising factorial) (a)n=a(a+1)⋯(a+n−1)
Hence
S=∞∑n=0zn(a)n+1=∞∑n=0zna(a+1)⋯(a+n)=∞∑n=0zna[(a+1)⋯(a+n)]=∞∑n=0zna(a+1)n=1a∞∑n=0(1)n(a+1)nzn(1)n=1a∞∑n=0(1)n(a+1)nznn!=1a1F1(1,a+1,z)
1F1(a,b,z) is the confluent hypergeometric function, also denoted M(a,b,z). This function satisfies the Kummer’s Transformation:
1F1(a,b,z)=ez1F1(b−a,b,−z)
Therefore
1a1F1(1,a+1,z)=eza1F1(a,a+1,−z)=eza∞∑n=0(a)n(a+1)n(−z)n=eza∞∑n=0a(a+1)⋯(a+n−1)(a+1)(a+2)⋯(a+n−1)(a+n)(−z)n=eza∞∑n=0a(a+1)⋯(a+n−1)(a+1)(a+2)⋯(a+n−1)(a+n)(−z)n=eza∞∑n=0a(a+n)(−z)n=ez∞∑n=0(−z)n(a+n)
Hence
∞∑n=0zn(a)n+1=ez∞∑n=0(−z)n(a+n)
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