Calcular $\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\ln n+\gamma)^2\right) $

Question

Calcular

$$\lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\ln n+\gamma)^2\right) $$ donde $\gamma$ - Euler constante.

Answer 1

Tenemos \begin{align} 2\sum_{k=1}^n \frac{H_k}{k} &= 2\sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{j=1}^n \sum_{k=j}^n \frac{1}{jk}, \text{ swapping the order of summation on the second sum}\\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=k}^n \frac{1}{jk}, \text{ changing variables on the second sum}\\ &= \sum_{k=1}^n \sum_{j=1}^n \frac{1}{jk} + \sum_{k=1}^n \frac{1}{k^2} \\ &= \left(\sum_{k=1}^n \frac{1}{k} \right)^2 + \sum_{k=1}^n \frac{1}{k^2} \\ &= H_n^2+ H^{(2)}_n. \\ \end{align}

Así \begin{align*} \lim_{n\to\infty} \left(\sum_{k=1}^n \frac{H_k}{k}-\frac{1}{2}(\log n+\gamma)^2\right) &= \lim_{n\to\infty} \frac{1}{2}\left(H_n^2+ H^{(2)}_n-(\log n+\gamma)^2\right) \\ &= \lim_{n\to\infty} \frac{1}{2}\left((\log n + \gamma)^2 + O(\log n/n) + H^{(2)}_n-(\log n+\gamma)^2\right) \\ &= \frac{1}{2}\lim_{n\to\infty} \left( H^{(2)}_n + O(\log n/n) \right) \\ &= \frac{1}{2}\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k^2}\\ &= \frac{\pi^2}{12}. \end{align*}