Leveled Up Limit

Since $H_n - \ln n \to \gamma$ , the Euler-Mascheroni constant, as $n \to \infty$ , we deduce that $\frac{H_n}{\ln n} \to 1$ as $n \to \infty$ . But then $\frac{H_{n^7}}{7\ln n} = \frac{H_{n^7}}{\ln n^7} \to 1$ as $n \to \infty$ , and so $\lim_{n \to \infty}(\ln n)^{-1}H_{n^7} \; = \; \boxed{7}$

In general for any $k\geq 1$ we can deduce that $\lim_{n\to \infty}\frac{1}{\ln n}\left(1+\frac{1}{2}+\frac{1}{2}+\cdots +\frac{1}{n^k}\right)=k$ and hence for $k=7$ we have required answer as $7$ .