Who's up to the challenge? 38

My method is to find a partial sum formula. Since partial sums have finitely many terms, we can break up the sum into three parts:

(1) First let's find a formula for $\displaystyle\sum_{n = 1}^{m} \dfrac{H_n}{n}$ . This sum expands to $\displaystyle\sum_{n = 1}^{m}\sum_{k = 1}^{n} \dfrac{1}{k}\dfrac{1}{n} = \sum_{k < n \le m} \dfrac{1}{k}\dfrac{1}{n} + \sum_{k = 1}^{m} \dfrac{1}{k^2}$ . This is close to the formula for $H_m^2$ , which is: $H_m^2 = \displaystyle\sum_{n = 1}^{m}\sum_{k = 1}^{m} \dfrac{1}{k}\dfrac{1}{n} = \sum_{k < n \le m} \dfrac{1}{k}\dfrac{1}{n} + \sum_{n < k \le m} \dfrac{1}{k}\dfrac{1}{n} + \sum_{k = 1}^{m} \dfrac{1}{k^2} = 2\left(\displaystyle\sum_{n = 1}^{m} \dfrac{H_n}{n}\right) - \displaystyle\sum_{k = 1}^{m} \dfrac{1}{k^2}$ Hence, $\displaystyle\sum_{n = 1}^{m} \dfrac{H_n}{n} = \dfrac{1}{2}\left(H_{m}^2 + \displaystyle\sum_{k=1}^{m} \dfrac{1}{k^2}\right)$

(2) To find the formula for $\displaystyle\sum_{n = 1}^{m} \left(- \dfrac{\ln(n)}{n}\right)$ , we need to use the generalized Stieltjes constant: $\gamma_{1}(m) = \lim_{r \to \infty} \left[ \sum_{k = m}^{r} \frac{\ln(k)}{k} - \frac{\ln^2(r)}{2} \right]$ Observe that: $\gamma_{1}(m+1) - \gamma_{1}(1) = \lim_{r \to \infty} \left[ \sum_{k = m+1}^{r} \frac{\ln(k)}{k} - \sum_{k = 1}^{r} \frac{\ln(k)}{k} \right] = -\sum_{k = 1}^{m} \dfrac{\ln(n)}{n}$ Therefore, $\displaystyle\sum_{n = 1}^{m} \left(- \dfrac{\ln(n)}{n}\right) = \gamma_{1}(m+1) - \gamma_1$

(3) The formula for $\displaystyle\sum_{n = 1}^{m} \left(- \dfrac{\gamma}{n}\right)$ is obviously $-\gamma H_m$

SO, putting all of this together, we get our total partial sum formula: $\sum_{n=1}^{m} \dfrac{H_n - \ln(n) - \gamma}{n} = \dfrac{1}{2}\left(H_{m}^2 + \displaystyle\sum_{k=1}^{m} \dfrac{1}{k^2}\right) + \gamma_{1}(m+1) - \gamma_1 -\gamma H_m$ $= \dfrac{(H_m - \gamma)^2}{2} + \gamma_{1}(m+1) - \dfrac{\gamma^2}{2} - \gamma_1 + \dfrac{1}{2}\sum_{k=1}^{m} \dfrac{1}{k^2}$

If we let $m$ approach infinity, the last three terms give us the desired result: $\boxed{-\dfrac{\gamma^2}{2} - \gamma_1 + \dfrac{\pi^2}{12}}$

All that remains is to prove that $\lim_{m \to\infty} \left(\dfrac{(H_m - \gamma)^2}{2} + \gamma_{1}(m+1)\right) = 0$ : $\lim_{m \to\infty} \left(\dfrac{(H_m - \gamma)^2}{2} + \gamma_{1}(m+1)\right)$ $= \lim_{m \to\infty} \left[\left(\dfrac{(H_m - \gamma)^2}{2} - \dfrac{\ln^2(m)}{2}\right) + \dfrac{\ln^2(m)}{2} + \lim_{r \to \infty} \left( \sum_{k = m+1}^{r} \frac{\ln(k)}{k} - \frac{\ln^2(r)}{2}\right) \right]$ $= \lim_{m \to\infty} \left(\dfrac{(H_m - \gamma)^2}{2} - \dfrac{\ln^2(m)}{2}\right) + \lim_{m \to\infty} \left(\dfrac{\ln^2(m)}{2} - \sum_{k = 1}^{m} \frac{\ln(k)}{k} \right) + \lim_{r \to \infty} \left( \sum_{k = 1}^{r} \frac{\ln(k)}{k} - \frac{\ln^2(r)}{2}\right)$ $= \lim_{m \to\infty} \left(\dfrac{(H_m - \gamma)^2}{2} - \dfrac{\ln^2(m)}{2}\right) - \gamma_1 + \gamma_1 = \dfrac{1}{2}\lim_{m \to\infty} \left((H_m - \gamma)^2 - \ln^2(m)\right)$ Now to prove that this limit is equal to zero, we need to know that $\lim_{m\to\infty} m\left(H_m - \ln(m) - \gamma\right) = \dfrac{1}{2}$ . I'll prove this at the end. $\dfrac{1}{2}\lim_{m \to\infty} \left((H_m - \gamma)^2 - \ln^2(m)\right) = \dfrac{1}{2}\lim_{m \to\infty} m\left(H_m - \ln(m) - \gamma\right)\left(\dfrac{H_m + \ln(m) - \gamma}{m}\right)$ $= \dfrac{1}{2} \cdot \dfrac{1}{2} \lim_{m \to \infty} \left(\dfrac{H_m + \ln(m) - \gamma}{m}\right)$ We can show using the Squeeze Theorem that $\lim_{m \to \infty} \left(\dfrac{H_m + \ln(m) - \gamma}{m}\right) = 0$ . Thus $\lim_{m \to\infty} \left(\dfrac{(H_m - \gamma)^2}{2} + \gamma_{1}(m+1)\right) = 0$ . $\square$

Now, as promised, I will prove that $\lim_{m\to\infty} m\left(H_m - \ln(m) - \gamma\right) = \dfrac{1}{2}$ :

$\lim_{m\to\infty} m\left(H_m - \ln(m) - \gamma\right) = \lim_{m\to\infty} \dfrac{\psi(m+1) - \ln(m)}{1/m}$ The antiderivative of the numerator is $\ln(\Gamma(m+1)) - m\ln(m) + m = \ln\left(m! \dfrac{e^m}{m^m}\right)$ . By Stirling's Approximation, this is asymptotic to $\ln(\sqrt{2\pi m})$ (and therefore approaches infinity). Meanwhile, the antiderivative of the denominator is $\ln(m)$ which also approaches infinity. Therefore, by L'Hopital's rule (in reverse), we have: $\lim_{m\to\infty} \dfrac{\psi(m+1) - \ln(m)}{1/m} = \lim_{m\to\infty} \dfrac{\ln\left(m! \dfrac{e^m}{m^m}\right)}{\ln(m)} = \lim_{m\to\infty} \log_m\left(\sqrt{2\pi m}\right) = \dfrac{1}{2}$