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+n−1)
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(2−4a)z((1+z)1−2a−(1−z)1−2a)
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)1−t−(1−i)1−t(2−2t)i
The derivative of the rising factorial is:
ddx(x)n=(x)n[ψ(n+x)−ψ(x)]
Hence, differentiating (1)
ddt∞∑n=0(t)2n(2)2n(−1)n=ddt[(1+i)1−t−(1−i)1−t(2−2t)i]
⟹∞∑n=0(t)2n[ψ(2n+t)−ψ(t)](2)2n(−1)n=ddt[(1+i)1−t−(1−i)1−t(2−2t)i]=2i((1−i)1−t−(1+i)1−t)(2−2t)2−i((1−i)1−tln(1−i)−(1+i)1−tln(1+i))2−2t
Taking the limit as t→1
∞∑n=0(1)2n[ψ(2n+1)+γ](2)2n(−1)n=limt→1[2i((1−i)1−t−(1+i)1−t)(2−2t)2⏟=0−i((1−i)1−tln(1−i)−(1+i)1−tln(1+i))2−2t⏟L'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
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