Does $\sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right) $ converge to 1? I'm trying to show convergence of the following:

$$
\lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right) 
$$

How shall I proceed?
Hint:  Draw a graph of $f(x) = \dfrac{1}{x}$ over $[1,n]$,and see that: $\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}&lt;\displaystyle \int_{1}^n \dfrac{1}{x}dx = \ln n &lt; 1+\dfrac{1}{2}+\cdots +\dfrac{1}{n-1}$
Since $\log(n)$ can be viewed as a number in the sum:

$$ \lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right) = \lim\limits_{n \to \infty} \frac{1}{\log(n)} \sum_{k=1}^{n} \left(\frac{1}{k }\right) = \lim\limits_{n \to \infty} \frac{\sum_{k=1}^{n} \left(\frac{1}{k }\right)}{\log(n)} \cdot  $$ since $$\lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k }\right)$$ diverges so look at this, you can apply L'Hopital Rule (that will be your work).