Singular series posted by @infseriesbot
Today we evaluate this series posted by @infseriesbot (∞∑n=0(2n)!n!3xn)2=e4x∞∑n=0(2n)!n!4x2n The proof will rely on some properties of the confluent hypergeometric function and the binomial coefficient
Proof:
We start with the left hand side. Recall that the Pocchhammer symbol or rising factorial can be expressed as quotient of two gamma functions (x)n=Γ(n+x)Γ(x) Hence (2n)!=Γ(2n+1)=(1)2n Pochhammer symbols also satisfy the duplication formula: (x)2n=4n(x2)n(1+x2)n Hence (2n)!=4n(12)n(1)n Note also that (1)n=n!. Hence, ∞∑n=0(2n)!n!3xn=∞∑n=04n(12)n(1)n(1)n(1)nxnn!=∞∑n=0(12)n(1)n(4x)nn!=1F1(12;1;4x) The right hand side is the confluent hypergometric function 1F1(a;b;x). This function is also denoted with M(a;b;x):
The confluent hypergometric function satisfies the Kummer's second theorem or second transformation: 1F1(a,2a,x)=ex/20F1(;a+12;x216) Therefore ∞∑n=0(2n)!n!3xn=1F1(12;1;4x)=e2x0F1(;1;x2) From the definition of 0F1: e2x0F1(;1;x2)=e2x∞∑n=0x2nn!2 Hence ∞∑n=0(2n)!n!3xn=e2x∞∑n=0x2nn!2 Squaring both sides (∞∑n=0(2n)!n!3xn)2=e4x(∞∑n=0x2nn!2)2 We can evaluate the right hand side with the Cauchy product: (∞∑n=0x2nn!2)2=∞∑n=0n∑j=01j!2(n−j)!2x2n=∞∑n=01n!2n∑j=0n!2j!2(n−j)!2x2n=∞∑n=0x2nn!2n∑j=0(nj)2 From the properties of the binomial coefficient we know that: J∑j=0(nj)2=(2nn)J≥n Therefore (∞∑n=0x2nn!2)2=∞∑n=0x2nn!2n∑j=0(nj)2=∞∑n=0x2nn!2(2nn)=∞∑n=0(2n)!n!4x2n Hence from (1),(2),(3) and (4) we can conclude: (∞∑n=0(2n)!n!3xn)2=e4x∞∑n=0(2n)!n!4x2n
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