limit of sequence involving ln and harmonic series

Question

I am trying to compute the following limit:

$$\lim_{n \to \infty}\frac{1}{\ln{n}}\sum_{k=1}^{\lfloor An \rfloor}\frac{1}{k}$$

where $A>0$.

I know the result for $A=1$, which is 1, and I tried to reproduce the proof using the Stolz-Cesaro theorem as in https://math.stackexchange.com/a/575713/854355, and i get that the limit is $\frac{1}{A}$, but it seems to me that it should be, for $A \geq 1$,

$$\lim_{n \to \infty}\frac{1}{\ln{n}}\sum_{k=1}^{\lfloor An \rfloor}\frac{1}{k} \geq \lim_{n \to \infty}\frac{1}{\ln{n}}\sum_{k=1}^{n}\frac{1}{k} = 1,$$

as $An \geq n$, and there are less terms in the sum.

Answer 1

The limit is always $1$. Indeed, let $H_n = \sum\limits_{k=1}^n \frac{1}{k}$.

$$\lim\limits_{n\to \infty} \frac{H_{\left\lfloor An\right\rfloor}}{\ln n} = \lim\limits_{n\to \infty} \frac{H_{\left\lfloor An\right\rfloor}}{\ln \left\lfloor An\right\rfloor} \times \frac{\ln \left\lfloor An\right\rfloor}{\ln n} = 1$$ since $\ln (An) = \ln A + \ln n$.