How to show this equation is true.
∞∑n=1Hnn2=2ζ(3)
where Hn=11+12+13+⋯+1n
Answer
It can be shown that
Hn=∫101−tn1−t dt.
Using this integral form of the Harmonic numbers the series in question becomes
S=∞∑n=11n2⋅∫101−tn1−t dt=∫10[ζ(2)−Li2(t)]dt1−t,
where Li2(x) is the dilogarithm. Now using the integral
∫Li2(t)1−t dt=2Li3(1−t)−2Li2(1−t) ln(1−t)−Li2(t) ln(1−t)−ln(t) ln2(1−t)
it is seen that, with the use of Lim(1)=ζ(m), Lim(0)=0, ln(1)=0,
S=ζ(2)[−ln(1−t)]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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