real analysis - Infinite Series $sumlimits_{n=1}^inftyleft(frac{H_n}nright)^2$

How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).

Answer

EDITED. Some simplifications were made.

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Here is a solution.

1. Basic facts on the dilogarithm. Let $\mathrm{Li}_{2}(z)$ be the dilogarithm function defined by

$$\operatorname{Li}_{2}(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n^{2}} = - \int_{0}^{z} \frac{\log(1-x)}{x} \, dx.$$

Here the branch cut of $\log$ is chosen to be $(-\infty, 0]$ so that $\operatorname{Li}_{2}$ defines a holomorphic function on the region $\Bbb{C} \setminus [1, \infty)$. Also, it is easy to check (by differentiating both sides) that the following identities hold

$$\begin{align*} \operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right) &= -\mathrm{Li}_{2}(z) - \tfrac{1}{2}\log^{2}(1-z); \quad z \notin [1, \infty) \tag{1} \\ \operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) &= \color{blue}{\boxed{\operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z)}} + \color{red}{\boxed{\log(-z)\log(1-z)}}; \quad z \notin [0, \infty) \tag{2} \end{align*}$$

Notice that in (2), the blue-colored part is holomorphic on $|z| < 1$ while the red-colored part induces the branch cut $[-1, 0]$.

2. A useful power series. Now let us consider the power series

$$f(z) = \sum_{n=0}^{\infty} \frac{H_n}{n} z^n.$$

Then $f(z)$ is automatically holomorphic inside the disc $|z| < 1$. Moreover, it is easy to check that

$$\sum_{n=1}^{\infty} H_{n} z^{n-1} = \frac{1}{z} \left( \sum_{n=1}^{\infty} \frac{z^{n}}{n} \right)\left( \sum_{n=0}^{\infty} z^{n}\right) = -\frac{\log(1-z)}{z(1-z)}.$$

thus integrating both sides, together with the identity $\text{(1)}$, we obtain the following representation of $f(z)$.

$$f(z) = \operatorname{Li}_{2}(z) + \tfrac{1}{2}\log^{2}(1-z) = -\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right). \tag{3}$$

3. Integral representation and the result. By the Parseval's identity, we have

$$\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} = \frac{1}{2\pi} \int_{0}^{2\pi} f(e^{it})f(e^{-it}) \, dt = \frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} f\left(\frac{1}{z}\right) \, dz \tag{4}$$

Since $\frac{1}{z}f(z)$ is holomorphic inside $|z| = 1$, the failure of holomorphy of the integrand stems from the branch cut of

$$\begin{align*} f\left(\tfrac{1}{z}\right) &= -\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) \\ &= -\color{blue}{\left( \operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z) \right)} - \color{red}{\log(-z)\log(1-z)}, \end{align*}$$

which is $[0, 1]$. To resolve this, we utilize the identity $\text{(2)}$. Note that the blue-colored portion does not contributes to the the integral $\text{(4)}$, since it remains holomorphic inside $|z| < 1$. That is, only the red-colored portion gives contribution to the integral. Consequently we have

$$\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &= -\frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} \color{red}{\log(-z)\log(1-z)} \, dz. \tag{5} \end{align*}$$

Since the integrand is holomorphic on $\Bbb{C} \setminus [0, \infty)$, we can utilize the keyhole contour wrapping around $[0, 1]$ to reduce $\text{(5)}$ to

$$\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &=-\frac{1}{2\pi i} \Bigg\{ \int_{0^{-}i}^{1+0^{-}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \\ &\qquad \qquad + \int_{1+0^{+}i}^{+0^{+}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \Bigg\} \\ &=-\frac{1}{2\pi i} \Bigg\{ \int_{0}^{1} \frac{f(x)(\log x + i\pi)\log(1-x)}{x} \, dx \\ &\qquad \qquad - \int_{0}^{1} \frac{f(x)(\log x - i\pi)\log(1-x)}{x} \, dx \Bigg\} \\ &=-\int_{0}^{1} \frac{f(x)\log(1-x)}{x} \, dx. \tag{5} \end{align*}$$

Plugging $\text{(3)}$ to the last integral and simplifying a little bit, we have

$$\begin{align*} \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}} &= - \int_{0}^{1} \frac{\operatorname{Li}_2(x)\log(1-x)}{x} \, dx - \frac{1}{2}\int_{0}^{1} \frac{\log^{3}(1-x)}{x} \, dx \\ &= \left[ \frac{1}{2}\operatorname{Li}_2(x)^2 \right]_0^1 - \frac{1}{2} \int_{0}^{1} \frac{\log^3 x}{1-x} \, dx \\ &= \frac{1}{2}\zeta(2)^{2} + \frac{1}{2} \Gamma(4)\zeta(4) \\ &= \frac{17\pi^{4}}{360} \end{align*}$$

as desired.