$\sum_{k=1}^n\frac1k\sim\ln n$ I would like to know how the author got  $\frac{\sum_{k=1}^{n}\frac{1}{k}}{\ln(n)} \longrightarrow 1$  from $ \sum_{k=1}^{n}\frac{1}{k} =\ln(n)+\lambda+o(1)$

$$\frac{\sum_{k=1}^{n}\frac{1}{k}}{ln(n)}=1+\frac{\lambda}{ \ln(n)} +o(\frac{1}{\ln(n)})$$
when n goes to infinity we got that 
$$ \frac{\sum_{k=1}^{n}\frac{1}{k}}{ln(n)}=1+o(0)$$
 A: Use the standard inequalities

$$\log n-\log 2 = \int_2^n {dt\over t}\le \sum_{j=2}^n{1\over k}\le \int_1^n{dt\over t} = \log n$$

Now divide everything by $\log n$ and take limits. The squeeze theorem gives the result.
A: You have $\sum_{k=1}^n\frac1k=\ln n+\gamma+o(1)$. So
$$\frac{\sum_{k=1}^n\frac1k}{\ln n}-1=\frac{\gamma+o(1)}{\ln n}.$$
In this fraction the numerator is bounded and the denominator tends to infinity, so the whole thing tends to zero.
A: 
I thought it might be instructive to present an approach that relies on elementary analysis only.


Let $a_n=\sum_{k=1}^n\frac 1k -\log(n)$.  Recalling that $\log(x)\ge \frac{x-1}{x}$ for all $x>0$, we have
$$a_{n+1}-a_n=\frac1{n+1}-\log\left(1+\frac1n\right)\le 0\tag 1$$
Therefore, $(1)$ establishes that the sequence $a_n$ is decreasing.  
We also see that 
$$a_n\ge 1+\int_2^n \frac1x\,dx-\log(n)=1-\log(2)\tag2$$
Therefore, $(2)$ establishes that $a_n$ is bounded below.  
Since $a_n$ is decreasing and bounded below, it converges.  
Now we denote by $\gamma$, the limit 
$$\gamma=\lim_{n\to \infty}\left(\sum_{k=1}^n\frac 1k -\log(n)\right)\tag 3$$
Then, using $(3)$ we can write 
$$\sum_{k=1}^n\frac1k =\log(n)+\gamma+o(1) \tag 4$$
where $o(1)$ is the "little o" notation.
From $(4)$ it is easy to see that 
$$\lim_{n\to \infty}\frac{\sum_{k=1}^n\frac1k}{\log(n)}=\lim_{n\to \infty}\left(1+\frac\gamma{\log(n)+}+\frac{o(1)}{\log(n)}\right)=1$$
as was to be shown!
A: $$\lim_{n\to +\infty}\frac{H_n}{\log n} = \lim_{n\to +\infty}\frac{\frac{1}{n+1}}{\log(n+1)-\log n} = \lim_{n\to +\infty}\frac{n}{n+1}\cdot\frac{1}{n\log\left(1+\frac{1}{n}\right)}=\color{red}{1} $$
simply follows from the Stoltz-Cesàro theorem.
As an alternative approach you may notice that $\frac{1}{n}=\log\left(1+\frac{1}{n}\right)+O\left(\frac{1}{n^2}\right)$ and that
$$ \sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\log(N+1) $$
because the LHS is a telescopic sum.
