Let $H_0=0$
\begin{align}
\sum_{n=1}^{k}\frac{H_n}{n} & = \sum_{n=1}^k\frac{\left(H_{n-1}+\frac{1}{n}\right)}{n}\\
& = \sum_{n=1}^k\frac{H_{n-1}}{n}+\frac{1}{n^2}\\
& = \sum_{n=1}^k\frac{H_{n-1}}{n}+\sum_{n=1}^k\frac{1}{n^2}\\
& = \sum_{n=1}^k\frac{H_{n-1}}{n}+H_{k,2}\\
& = \sum_{n=1}^{k-1}\frac{(H_k-H_n)}{n}+H_{k,2}\\
& = \sum_{n=1}^{k-1}\frac{H_k}{n}-\sum_{n=1}^{k-1}\frac{H_n}{n}+H_{k,2}\\
& = H_kH_{k-1}-\sum_{n=1}^{k}\frac{H_n}{n}+\frac{H_k}{k}+H_{k,2}\\
\therefore 2\sum_{n=1}^{k}\frac{H_n}{n} & = H_kH_{k-1}+\frac{H_k}{k}+H_{k,2}\\

\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n} & = \lim_{k\to \infty}\frac{1}{2}\left(H_kH_{k-1}+\frac{H_k}{k}+H_{k,2}\right)\\
& = \frac{1}{2}\left(\lim_{k\to \infty}\left(H_{k-1}^2+\frac{H_{k-1}}{k}\right)+\lim_{k\to \infty}\frac{H_k}{k}+\lim_{k\to \infty}H_{k,2}\right)\\
& = \frac{1}{2}\left(\lim_{k\to \infty}\left((\gamma+\ln (k-1))^2+\frac{\gamma+\ln (k-1)}{k}\right)+\lim_{k\to \infty}\frac{\gamma+\ln k}{k}+\zeta(2)\right)\\
& = \frac{1}{2}\left(\lim_{k\to \infty}(\gamma+\ln (k-1))^2+0+0+\zeta(2)\right)\\

Question: $f: \Omega \to \Bbb C$ is holomorphic s.t $f(\Omega) \subset \{\alpha z_0 : \alpha \in \Bbb R \}$ for some $z_0 \in \Bbb C$ .
i need to prove that $f$ is constant.
if $z=1$ , then if $f = u + v i $ , we have $u(z) + iv(z) = \alpha$ so by the Cauchy–Riemann equations we get that $v' = 0$ which implies $u' = 0$ so $f$ is constant .

in the general case, we have $u(z) = \alpha(z) x_0 , v(z) = \alpha(z) y_0$ where $z_0 = x_0 + y_0 i$. im not sure how to apply Cauchy–Riemann equations here. someone can help ?