asymptotics of sum I wanna find asymptotic of sum below
$$\sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k$$
assume I know asymptotic of this sum (I can be wrong):
$$\sum\limits_{k=1}^{n}\frac{1}{k}(1 - \frac{1}{n^2})^k \sim c\ln{n}$$
So I use Stolz–Cesàro theorem and wanna show that
$$\sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k \sim c\ln{n}$$
where 
$$x_n = \sum\limits_{k=1}^{[\sqrt{n}]}\frac{1}{k}(1 - \frac{1}{n})^k$$
$$x_n - x_{n-1} = \frac{1}{\sqrt{n}}(1 - \frac{1}{n})^{\sqrt{n}}$$
$$y_n - y_{n-1} = \ln(n) - \ln(n-1)$$
but
$$
\lim_{n \to \infty} 
\frac{\frac{1}{\sqrt{n}}(1 - \frac{1}{n})^{\sqrt{n}}}
{\ln(n) - \ln(n-1)} = \infty
$$
What I'm doing wrong?

Another method you could try is rewriting your sum as
$$
\sum_{k=1}^{\lfloor\sqrt{n}\rfloor} \frac{1}{k} - \sum_{k=1}^{\lfloor\sqrt{n}\rfloor} \frac{1}{k} \left[1-\left(1-\frac{1}{n}\right)^k\right],
$$
then using Bernoulli's inequality to show that the second sum is $O(n^{-1/2})$.
