Tuesday, 2 January 2018

summation - How to show suminftyn=1fracHnn2=2zeta(3)?




How to show this equation is true.



n=1Hnn2=2ζ(3)




where Hn=11+12+13++1n


Answer



It can be shown that
Hn=101tn1t dt.
Using this integral form of the Harmonic numbers the series in question becomes
S=n=11n2101tn1t dt=10[ζ(2)Li2(t)]dt1t,
where Li2(x) is the dilogarithm. Now using the integral
Li2(t)1t dt=2Li3(1t)2Li2(1t) ln(1t)Li2(t) ln(1t)ln(t) ln2(1t)
it is seen that, with the use of Lim(1)=ζ(m), Lim(0)=0, ln(1)=0,
S=ζ(2)[ln(1t)]10+ζ(2)ln(0)+2Li3(1)=ζ(2)ln(o)+ζ(2)ln(0)+2Li3(1)=2Li3(1)
which yields
n=1Hnn2=2ζ(3).


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