Processing math: 100%

Sunday, December 12, 2021

Generalized hypergeometric functions VII

Nice series involving Harmonic numbers

Nice series involving the harmonic number H2n


Totday we show the proof of this nice series posted by @BenriBot_ccbs n=1(1)nH2n2n+1=πln(2)8
The proof relies on some properties of the rising factorial and the Gauss hypergometric function 2F1.

Proof

Consider the following series: S=n=0(t)2n(2)2nxn
where (x)n=x(1+x)(x+n1)
is the rising factorial or Pochhammer polynomial

The rising factorial satisfies the duplication formula: (x)2n=4n(1x)n(1+x2)n
Hence S=n=0(t)2n(2)2nxn=n=04n(t2)n(1+t2)n4n(1)n(32)nxn=n=0(t2)n(1+t2)n(32)nxnn!(1)n=n!=2F1(t2,t+12;32,x)
where 2F1 is the Gauss hypergeometric function. 2F1 satisfies: 2F2(a,12+a;32;z2)=1(24a)z((1+z)12a(1z)12a)
If we put a=t2 and z2=i2=1 n=0(t)2n(2)2n(1)n=2F1(t2,t+12;32;1)=(1+i)1t(1i)1t(22t)i
The derivative of the rising factorial is: ddx(x)n=(x)n[ψ(n+x)ψ(x)]
Hence, differentiating (1) ddtn=0(t)2n(2)2n(1)n=ddt[(1+i)1t(1i)1t(22t)i]
n=0(t)2n[ψ(2n+t)ψ(t)](2)2n(1)n=ddt[(1+i)1t(1i)1t(22t)i]=2i((1i)1t(1+i)1t)(22t)2i((1i)1tln(1i)(1+i)1tln(1+i))22t
Taking the limit as t1 n=0(1)2n[ψ(2n+1)+γ](2)2n(1)n=limt1[2i((1i)1t(1+i)1t)(22t)2=0i((1i)1tln(1i)(1+i)1tln(1+i))22tL'Hôpital's rule ]=πln(2)8
Using the recursion formula (n+x)(x)n=x(x+1)n
Hence (2n+1)=(2)2n(1)2n
Therefore n=0[ψ(2n+1)+γ]2n+1(1)n=πln(2)8
Finally, using the relation of the Harmonic number to the digamma function: Hx=ψ(x+1)+γ
and H0=0

We can conclude n=1(1)nH2n2n+1=πln(2)8

No comments:

Post a Comment

Series of the day

Series involving the digamma and the zeta functions The sum 1(n+1)pnq ...