Does $\sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right) $ converge to 1?

Question

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?

Answer 1

Hint: Draw a graph of $f(x) = \dfrac{1}{x}$ over $[1,n]$,and see that: $\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}<\displaystyle \int_{1}^n \dfrac{1}{x}dx = \ln n < 1+\dfrac{1}{2}+\cdots +\dfrac{1}{n-1}$

Answer 2

$$\frac1{\log n}\sum^n_{k=1}\frac1k=\frac{H_n}{\log n}\sim\frac{\gamma+\log n}{\log n}\to 1$$ as $n\to\infty$.

Answer 3

Hint

$$S_n=\sum_{k=1}^{n} \frac{1}{k \, \log(n)}=\frac{1}{ \log(n)}\sum_{k=1}^{n} \frac{1}{k}=\frac{H_n}{ \log(n)}$$ where appears the harmonic number.

Use the asymptotics $$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$

Answer 4

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).