# finding a harmonic sum using residues/complex analysis

Evaluate:

$$S = \sum_{n=1}^{\infty} \frac{H_n}{n^2}$$

Using complex analysis.

I just needs hints, I have no attempts,

but I believe is has to do with residues.

• Have you seen any similar problems done with residues? – Gerry Myerson Jan 16 '15 at 19:57
• I have but not with $H_n$ – Amad27 Jan 16 '15 at 19:57
• what is the meaning of $H_{n}$? – Mark Hubenthal Jan 17 '15 at 2:03
• @Mark, probably $H_n=\sum_1^n(1/k)$. – Gerry Myerson Jan 17 '15 at 3:40

By using a lemma you already asked to prove, if we take $f(z)=(\gamma+\psi(-z))^2$, the poles of $f(z)$ occur in $z=0,1,2,\ldots$ and for any $n\in\mathbb{N}^*$ we have: $$\operatorname{Res}\left(f(z),z=n\right) = 2H_n,$$ so: $$\sum_{n=1}^{+\infty}\frac{H_n}{n^2}=-\operatorname{Res}\left(\frac{f(z)}{z^2},z=0\right)=-\operatorname{Res}\left(\left(\frac{\gamma+\psi(-z)}{z}\right)^2,z=0\right).\tag{1}$$ On the other hand, since in a punctured neighbourhood of $z=0$: $$\gamma+\psi(-z) = \frac{1}{z}-\zeta(2) z-\zeta(3)z^2-\zeta(4)z^4+\ldots \tag{2}$$ it happens that: $$\left(\frac{\gamma+\psi(-z)}{z}\right)^2=\frac{1}{z^4}-\frac{2\zeta(2)}{z^2}-\frac{2\zeta(3)}{z}+\frac{\zeta(4)}{2}+\ldots\tag{3}$$ so, by $(1)$, $$\sum_{n=1}^{+\infty}\frac{H_n}{n^2}=\color{red}{2\,\zeta(3)}.\tag{4}$$
• Are you using the zeta definition of $\psi(-z) + \gamma$ in $(2)$ ?? Where can I see it? – Amad27 Jan 17 '15 at 14:10
• From en.wikipedia.org/wiki/Digamma_function#Taylor_series $$\psi(z+1) + \gamma = -\sum_{n=1}^{\infty} \zeta(n+1)(-z)^n$$ $$\psi(-z) + \gamma = -\sum_{n=1}^{\infty} \zeta(n+1)(z + 1)^n$$ this does give your series. – Amad27 Jan 17 '15 at 14:12
• @Amad27, you are asking the same things over and over. The Laurent series in $(2)$ arises from considering the logarithmic derivative of the Weierstrass product for the $\Gamma$ function, or the integral representation of the $\psi$ function. Anyway, I remember you used relation $(2)$ somewhere, so you should be aware of its proof. – Jack D'Aurizio Jan 17 '15 at 14:15
• And remember that $\psi(z+1)=\frac{1}{z}+\psi(z)$. – Jack D'Aurizio Jan 17 '15 at 14:16